Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

A plane is uniquely determined by a point and a normal vector perpendicular to its surface. Since the required plane is perpendicular to two other planes, its normal vector must be perpendicular to the normal vectors of both given planes, which means it can be calculated as the cross product of those two vectors. The planes $2x+4y+5z=8$ and $3x-2y+3z=5$ have normal vectors $\vec{n}_1 = (2, 4, 5)$ and $\vec{n}_2 = (3, -2, 3)$, respectively.

We determine the normal vector $\vec{n}$ by calculating the cross product $\vec{n}_1 \times \vec{n}_2 = (4(3) - 5(-2), 5(3) - 2(3), 2(-2) - 4(3))$, which simplifies to the vector $(22, 9, -16)$. Utilizing the point-normal form for a plane passing through the point $(-2, 3, 5)$, we set up the equation $22(x+2) + 9(y-3) - 16(z-5) = 0$. Expanding this expression leads to $22x + 44 + 9y - 27 - 16z + 80 = 0$, which simplifies to $22x + 9y - 16z + 97 = 0$. By comparing this result with the given form $\alpha x + \beta y + \gamma z + 97 = 0$, we identify $\alpha = 22$, $\beta = 9$, and $\gamma = -16$. Calculating the sum of these coefficients yields $\alpha + \beta + \gamma = 22 + 9 - 16 = 15$.

Reflecting a point across a plane involves identifying a line that passes through the given point and is perpendicular to the plane, then finding the intersection point, which serves as the midpoint between the original point and its image. For the point $P(2,-1,3)$ and the plane $x+2y-z=0$, the normal vector of the plane, which defines the direction of the perpendicular line, is $(1, 2, -1)$. We can express any point on this line using a parameter $k$ as $(2+k, -1+2k, 3-k)$. Substituting these coordinates into the plane equation $(2+k) + 2(-1+2k) - (3-k) = 0$ results in $6k - 3 = 0$, confirming that the intersection occurs at $k=1/2$.

Evaluating the line coordinates at $k=1/2$ gives the midpoint $(2.5, 0, 2.5)$. Given that this point is the average of $P(2,-1,3)$ and its image $Q(\alpha, \beta, \gamma)$, we set $\frac{\alpha+2}{2}=2.5$, $\frac{\beta-1}{2}=0$, and $\frac{\gamma+3}{2}=2.5$, which yields the coordinates for the image $Q(3,1,2)$.

Finally, calculating the perpendicular distance from point $Q$ to the plane $3x+2y+z+29=0$ is achieved using the formula $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Substituting the coordinates of $Q$ into this formula gives $d = \frac{|3(3)+2(1)+1(2)+29|}{\sqrt{3^{2+}2^{2+}1^2}}$, which simplifies to $\frac{42}{\sqrt{14}}$. Rationalizing this value results in a distance of $3\sqrt{14}$.

Any plane passing through the intersection of the surfaces defined by $x-2y-z-5=0$ and $x+y+3z-5=0$ can be represented using the family of planes equation $(x-2y-z-5) + \lambda(x+y+3z-5) = 0$. Rearranging this into standard form gives $(1+\lambda)x + (\lambda-2)y + (3\lambda-1)z = 5(1+\lambda)$. Since the plane must be parallel to the line defined by the intersection of $x+y+2z-7=0$ and $2x+3y+z-2=0$, the normal vector of our plane, $(1+\lambda, \lambda-2, 3\lambda-1)$, must be orthogonal to the direction vector of this second line. The direction vector of the line is the cross product of the normal vectors of the two planes defining it, $(1, 1, 2)$ and $(2, 3, 1)$, which yields $\mathbf{i}(1-6) - \mathbf{j}(1-4) + \mathbf{k}(3-2) = (-5, 3, 1)$.

Setting the dot product of the plane's normal vector and the line's direction vector to zero enforces the condition of parallelism: $-5(1+\lambda) + 3(\lambda-2) + 1(3\lambda-1) = 0$. Expanding this equation leads to $-5 - 5\lambda + 3\lambda - 6 + 3\lambda - 1 = 0$, which simplifies to $\lambda - 12 = 0$, identifying the parameter $\lambda=12$. Substituting $\lambda=12$ into the family of planes equation results in $13x+10y+35z=65$, allowing us to determine the constants as $a=13$, $b=10$, and $c=35$. The perpendicular distance from the point $(13, 10, 35)$ to the plane $2x+2y-z+16=0$ is determined using the distance formula:

$d = \frac{|2(13) + 2(10) - 35 + 16|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{|26 + 20 - 35 + 16|}{\sqrt{9}} = \frac{27}{3} = 9$

When a line is parallel to a plane, the direction vector of the line must be perpendicular to the normal vector of the plane, implying that their dot product equals zero. Given the plane $P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ and the line $L$ with the direction vector $(2, 3, 5)$, we establish the relationship $8(2) + \alpha_1(3) + \alpha_2(5) = 0$, which simplifies to the equation $3\alpha_1 + 5\alpha_2 = -16$. Additionally, the $y$-intercept of $1$ means the plane passes through the point $(0, 1, 0)$. Substituting these coordinates into the plane equation yields $8(0) + \alpha_1(1) + \alpha_2(0) + 12 = 0$, directly giving us $\alpha_1 = -12$. Placing this value back into our parallel condition, we find $3(-12) + 5\alpha_2 = -16$, which leads to $5\alpha_2 = 20$, or $\alpha_2 = 4$.

With these constants determined, the equation for the plane becomes $8x - 12y + 4z + 12 = 0$, which can be simplified by dividing by $4$ to get $2x - 3y + z + 3 = 0$. Because the line $L$ is parallel to the plane, the distance between them is the constant perpendicular distance from any point on the line to the plane. Using the point $(-2, 3, -4)$ from the line equation, we apply the perpendicular distance formula $\frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$, resulting in $\frac{|2(-2) - 3(3) + 1(-4) + 3|}{\sqrt{2^2 + (-3)^2 + 1^2}}$. This calculation simplifies to $\frac{|-4 - 9 - 4 + 3|}{\sqrt{4 + 9 + 1}}$, which evaluates to $\frac{|-14|}{\sqrt{14}}$, ultimately resulting in a distance of $\sqrt{14}$.

The shortest distance between two skew lines is the length of the segment perpendicular to both, calculated using the scalar projection of the vector connecting any point on the first line to any point on the second line onto the cross product of their direction vectors.

The first line passes through $(2, 6, 2)$ with the direction vector $\mathbf{v_1} = \langle 2, 1, -2 \rangle$, which is the normal vector to the plane $2x+y-2z=10$. The second line $\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$ has the direction vector $\mathbf{v_2} = \langle 2, -3, 2 \rangle$ and passes through the point $(-1, -4, 0)$. The distance $d$ is given by the formula
$d = \frac{|(\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{v_1} \times \mathbf{v_2})|}{|\mathbf{v_1} \times \mathbf{v_2}|}$
where $\mathbf{r_1} = (2, 6, 2)$ and $\mathbf{r_2} = (-1, -4, 0)$.

The vector connecting these two points is $\mathbf{r_2} - \mathbf{r_1} = \langle -3, -10, -2 \rangle$. The cross product of the direction vectors results in $\mathbf{v_1} \times \mathbf{v_2} = \langle -4, -8, -8 \rangle$, which has a magnitude of 12. Computing the absolute value of the dot product between the connecting vector and the cross product vector gives $|12 + 80 + 16|$, which simplifies to 108. Dividing 108 by the magnitude 12 yields a final shortest distance of 9.

The normal vector to the plane, determined by the cross product of two vectors formed by the given points, is $3\hat{i} - 4\hat{j} + 12\hat{k}$. Using this normal and one of the given points, the plane's equation is defined as $3x - 4y + 12z = 3$ The distance from point $A(3,4,\alpha)$ to this plane is given by the formula $\frac{|3(3) - 4(4) + 12\alpha - 3|}{13} = 2$. Solving this absolute value equation for a natural number $\alpha$ yields $12\alpha - 10 = 26$, which results \in $\alpha = 3$. With $\alpha$ determined, the coordinates of point $B$ become $(2,3,a)$, and applying the distance formula again gives $\frac{|3(2) - 4(3) + 12a - 3|}{13} = 3$ Solving $|12a - 9| = 39$ leads directly to the positive value $a = 4$.

The reflection of a point across a plane depends on the normal vector of the plane, as the segment connecting the point to its image is parallel to this normal and is bisected by the plane. For the point $P(2, 3, 5)$ and the plane defined by $2x + y - 3z - 6 = 0$, we relate the coordinates of the image $(\alpha, \beta, \gamma)$ to the normal components $(2, 1, -3)$ through the constant ratio derived from the plane equation:
$\frac{\alpha - 2}{2} = \frac{\beta - 3}{1} = \frac{\gamma - 5}{-3} = -2 \left( \frac{2(2) + 3 - 3(5) - 6}{2^2 + 1^2 + (-3)^2} \right)$
Calculating the terms within the parentheses, the numerator becomes $4 + 3 - 15 - 6 = -14$ and the denominator becomes $4 + 1 + 9 = 14$. This yields a common ratio of $-2 \left( \frac{-14}{14} \right)$, which simplifies to $2$. By setting each individual coordinate expression equal to $2$, we obtain $\alpha - 2 = 4$, $\beta - 3 = 2$, and $\gamma - 5 = -6$, leading to the values $\alpha = 6$, $\beta = 5$, and $\gamma = -1$. The required sum $\alpha + \beta + \gamma$ is therefore $6 + 5 - 1 = 10$.

The shortest distance between two skew lines defined by position vectors $\vec{a_1}$ and $\vec{a_2}$ and direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by the formula:
$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$
The lines pass through points $(-2, 0, 5)$ and $(4, 1, -3)$ with direction vectors $\vec{b_1} = \langle 1, -2, 2 \rangle$ and $\vec{b_2} = \langle 1, 2, 0 \rangle$. The difference between the position vectors is $\vec{a_2} - \vec{a_1} = \langle 6, 1, -8 \rangle$.

Evaluating the cross product of the direction vectors involves the determinant:
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{vmatrix} = -4\hat{i} + 2\hat{j} + 4\hat{k}$
The magnitude of this resulting vector is $\sqrt{(-4)^2 + 2^2 + 4^2} = 6$. The dot product of the difference vector $\langle 6, 1, -8 \rangle$ and this cross product vector is $(6)(-4) + (1)(2) + (-8)(4) = -54$. Dividing the absolute value of this result by the calculated magnitude yields $\frac{|-54|}{6} = 9$.

To determine the distance from the origin to the foot of the perpendicular, we first find the equation of the plane passing through the points $(-1, -2, -3)$, $(9, 3, 4)$, and $(9, -2, 1)$. By forming two vectors within the plane using these points, we calculate their cross product to find the normal vector to the plane, which results in $(2, 3, -5)$. Using the point-normal form of a plane equation with one of the given points, we obtain $2(x+1) + 3(y+2) - 5(z+3) = 0$, simplifying to the standard form $2x + 3y - 5z - 7 = 0$.

The foot of the perpendicular $Q(\alpha, \beta, \gamma)$ dropped from the point $P(3, -2, -9)$ to this plane satisfies the relation $\frac{x-3}{2} = \frac{y+2}{3} = \frac{z+9}{-5} = -\frac{2(3) + 3(-2) - 5(-9) - 7}{2^2 + 3^2 + (-5)^2}$. Simplifying the right side of this equation yields $-\frac{38}{38}$, which equals $-1$. Setting each parametric ratio equal to $-1$, we determine the coordinates of $Q$ by calculating $x = 3 + 2(-1) = 1$, $y = -2 + 3(-1) = -5$, and $z = -9 - 5(-1) = -4$, giving us the point $Q(1, -5, -4)$. The distance from the origin to this point is simply the square root of the sum of the squares of its coordinates, $\sqrt{1^2 + (-5)^2 + (-4)^2}$, which evaluates to $\sqrt{1 + 25 + 16} = \sqrt{42}$.

Coplanar lines must lie within the same plane, implying that the line $l_1$ resides within a plane belonging to the family defined by $(3x+2y+z-2) + \mu(x-3y+2z-13)=0$. For $l_1$ to lie on such a plane, its direction vector $\langle3, 1, -2\rangle$ must be perpendicular to the plane's normal vector, a condition satisfied by the equation $3(3+\mu)+1(2-3\mu)-2(1+2\mu)=0$, which simplifies to $9-4\mu=0$ and yields $\mu = \frac{9}{4}$. By ensuring that a point from $l_1$, such as $(-5, -4, \alpha)$, satisfies the specific plane equation $4(3x+2y+z-2) + 9(x-3y+2z-13)=0$, we find that $\alpha=7$.

With $l_1$ fully defined as passing through $(-5, -4, 7)$ with direction $\langle3, 1, -2\rangle$, any point $P$ on this line is given by $(3\lambda-5, \lambda-4, -2\lambda+7)$. Because $P$ is the point on $l_1$ nearest to $Q(-4, -3, 2)$, the vector $\vec{PQ} = \langle 3\lambda-1, \lambda-1, -2\lambda+5 \rangle$ must be orthogonal to the direction of $l_1$. The dot product $\langle 3\lambda-1, \lambda-1, -2\lambda+5 \rangle \cdot \langle3, 1, -2\rangle = 0$ leads to the equation $3(3\lambda-1)+1(\lambda-1)-2(-2\lambda+5)=0$, resulting in $14\lambda - 14 = 0$ and $\lambda = 1$. Consequently, the point $P$ is $(-2, -3, 5)$, and the sum of the absolute values of its coordinates is $|-2|+|-3|+|5| = 10$.

The first step is to establish the equation for the second plane, $P_2$. Given the three points $(2, -1, 0)$, $(2, 0, -1)$, and $(5, 1, 1)$, we define two vectors within the plane: $(0, 1, -1)$ and $(3, 2, 1)$. The normal vector to $P_2$ is the cross product of these two vectors, which results in $(3, -3, -3)$. Simplifying this to $(1, -1, -1)$ provides the normal, so the equation $1(x-2) - 1(y+1) - 1(z-0) = 0$ simplifies to $x - y - z = 3$.

Next, we identify the line of intersection between $P_1: 3x - y - 7z = 11$ and $P_2: x - y - z = 3$. The direction vector of this line is the cross product of the normal vectors of the two planes, $\vec{n}_1 = (3, -1, -7)$ and $\vec{n}_2 = (1, -1, -1)$. Calculating this cross product yields $(-6, -4, -2)$, which we can treat as $(6, 4, 2)$ for simplicity. By setting $z=0$ in the equations of both planes, we solve the system $3x - y = 11$ and $x - y = 3$ to find a point on the line, which gives $x=4$ and $y=1$. Thus, the line of intersection is expressed in parametric form as $\frac{x-4}{6} = \frac{y-1}{4} = \frac{z-0}{2} = r$, giving any point on the line as $(\alpha, \beta, \gamma) = (6r+4, 4r+1, 2r)$.

To find the foot of the perpendicular $(\alpha, \beta, \gamma)$ from the point $(7, 4, -1)$ to this line, we note that the vector connecting $(7, 4, -1)$ to the parametric point must be orthogonal to the line's direction vector $(6, 4, 2)$. Setting the dot product of these vectors to zero, we calculate $6(6r+4-7) + 4(4r+1-4) + 2(2r+1) = 0$. Expanding this gives $6(6r-3) + 4(4r-3) + 2(2r+1) = 0$, which simplifies to $36r - 18 + 16r - 12 + 4r + 2 = 0$, or $56r = 28$. This yields $r = \frac{1}{2}$. Substituting this value back into the expressions for $\alpha, \beta,$ and $\gamma$ gives $\alpha = 7, \beta = 3,$ and $\gamma = 1$. The sum $\alpha + \beta + \gamma$ is therefore $7 + 3 + 1 = 11$.

Calculating the distance from a point to a plane requires first determining the unique coordinates of the point that satisfies the given system of linear equations. By performing row operations on the system, such as subtracting the first equation $-x + 2y - 9z = 7$ from the second equation $-x + 3y + 7z = 9$, we derive the expression $y + 16z = 2$. Similarly, subtracting twice the first equation from the third equation $-2x + y + 5z = 8$ results in the linear combination $-3y + 23z = -6$. Solving this reduced pair of equations yields $z = 0$, which subsequently gives $y = 2$ and $x = -3$. The unique solution point is therefore $(\alpha, \beta, \gamma) = (-3, 2, 0)$.

The parameter $\lambda$ is determined by substituting these coordinates into the final given equation $-3x + y + 13z = \lambda$, which leads to $-3(-3) + 2 + 13(0) = 9 + 2 = 11$. With the point identified as $(-3, 2, 0)$ and the plane defined by $2x - 2y + z - 11 = 0$, the perpendicular distance is calculated using the formula
$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$
Substituting the identified values into this formula gives $\frac{|2(-3) - 2(2) + 1(0) - 11|}{\sqrt{2^2 + (-2)^2 + 1^2}}$, which simplifies to $\frac{|-6 - 4 - 11|}{\sqrt{9}}$. This calculation results in a final distance of $\frac{21}{3}$, which equals $7$.

The first line, represented by $\frac{x}{1} = \frac{6-y}{2} = \frac{z+8}{5} = \lambda$, yields the parametric coordinates $(\lambda, 6-2\lambda, 5\lambda-8)$. By equating these to the second line $\frac{x-5}{4} = \frac{y-7}{3} = \frac{z+2}{1} = \mu$, we establish the equations $\lambda = 4\mu+5$ and $6-2\lambda = 3\mu+7$. Solving this system results in $\lambda = 1$ and $\mu = -1$, yielding the intersection point $A$ at $(1, 4, -3)$.

Following a similar process for the intersection of the first line with the third line $\frac{x+3}{6} = \frac{3-y}{3} = \frac{z-6}{1} = \gamma$, we set $\lambda = 6\gamma-3$ and $6-2\lambda = 3-3\gamma$. Solving for $\gamma$ gives $\gamma=1$, which corresponds to $\lambda=3$ and identifies the intersection point $B$ at $(3, 0, 7)$. The midpoint of the segment $AB$ is the average of these coordinates, $(2, 2, 2)$.

The perpendicular distance from this midpoint $(2, 2, 2)$ to the plane $2x-2y+z=14$ is found using the distance formula:
$d = \frac{|2x_0-2y_0+z_0-14|}{\sqrt{2^{2+}(-2)^{2+}1^2}}$
Substituting the midpoint coordinates gives $\frac{|2(2)-2(2)+2-14|}{\sqrt{4+4+1}}$, which simplifies to $\frac{|-12|}{\sqrt{9}}$, resulting in a final distance of 4.

Two planes are perpendicular to each other when the dot product of their normal vectors is equal to zero. Given the planes $2x + ky - 5z = 1$ and $3kx - ky + z = 5$, the normal vectors are $(2, k, -5)$ and $(3k, -k, 1)$, respectively. The condition of perpendicularity requires that their dot product is zero, which gives $(2)(3k) + (k)(-k) + (-5)(1) = 0$. This simplifies to the quadratic equation $6k - k^2 - 5 = 0$, or $k^2 - 6k + 5 = 0$. Factoring this expression yields $(k-1)(k-5) = 0$, leading to the solutions $k=1$ or $k=5$. Since the problem specifies that $k < 3$, we must conclude that $k=1$.

Substituting $k=1$ into the given plane equations results in $2x + y - 5z = 1$ and $3x - y + z = 5$. The family of planes passing through the line of intersection of these two planes is represented by the equation $(2x + y - 5z - 1) + \lambda(3x - y + z - 5) = 0$. Because the plane $P$ intercepts a unit length of 1 on the positive $x$-axis, it must pass through the coordinate point $(1, 0, 0)$. Substituting these values into the family equation gives $(2(1) + 0 - 5(0) - 1) + \lambda(3(1) - 0 + 0 - 5) = 0$, which simplifies to $1 + \lambda(-2) = 0$, or $\lambda = \frac{1}{2}$.

With $\lambda = \frac{1}{2}$, the equation for the plane becomes:
$(2x + y - 5z - 1) + \frac{1}{2}(3x - y + z - 5) = 0$
Multiplying the entire equation by 2 eliminates the fraction, resulting in $4x + 2y - 10z - 2 + 3x - y + z - 5 = 0$. Combining the like terms leads to the plane equation $7x + y - 9z = 7$. To find the intercept made by this plane on the $y$-axis, we set $x=0$ and $z=0$, which yields $y = 7$.

The direction vector of a line defined by the intersection of two planes is obtained by calculating the cross product of their respective normal vectors. For the planes $lx - y + 3(1-l)z = 1$ and $x + 2y - z = 2$, the normal vectors are $(l, -1, 3(1-l))$ and $(1, 2, -1)$, which yield the direction vector $\vec{n} = (6l - 5)\hat{i} + (3 - 2l)\hat{j} + (2l + 1)\hat{k}$ through the cross product. Since this line lies within the plane $3x - 8y + 7z = 4$, its direction vector must be orthogonal to the normal vector of that plane, $(3, -8, 7)$. By setting the dot product of these two vectors to zero, we establish the equation $3(6l - 5) - 8(3 - 2l) + 7(2l + 1) = 0$, which simplifies to $18l - 15 - 24 + 16l + 14l + 7 = 0$, resulting in $48l = 32$ or $l = 2/3$.

Substituting $l = 2/3$ back into the expression for the direction vector gives $\vec{n} = (6(2/3) - 5, 3 - 2(2/3), 2(2/3) + 1)$, which simplifies to $(-1, 5/3, 7/3)$. This vector is parallel to $(-3, 5, 7)$, which simplifies the subsequent calculations. The cosine of the angle $\theta$ between this line and the $y$-axis is given by the ratio of the absolute value of the dot product of the direction vector with the $y$-axis unit vector, $(0, 1, 0)$, to the product of their magnitudes, yielding $\cos \theta = |( -3, 5, 7) \cdot (0, 1, 0)| / (\sqrt{(-3)^2 + 5^2 + 7^2} \cdot 1) = 5 / \sqrt{9 + 25 + 49} = 5/\sqrt{83}$. Consequently, the value of $415 \cos^2 \theta$ is calculated as $415 \times (25/83)$, which equals $5 \times 25 = 125$.

A plane is uniquely determined by a point it passes through and a vector normal to its surface. The normal vector of a plane containing two lines is the cross product of their direction vectors, while the normal vector of a plane perpendicular to another plane is perpendicular to the normal vector of that second plane.

To find the normal vector $\vec{n}_Q$ for the plane containing the lines $\frac{x}{2}= \frac{y}{3}= \frac{z}{5}$ and $\frac{x}{3}= \frac{y}{7}= \frac{z}{8}$, we calculate the cross product of their directions $\langle 2, 3, 5 \rangle$ and $\langle 3, 7, 8 \rangle$:
$\vec{n}_Q = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 5 \\ 3 & 7 & 8 \end{vmatrix} = \mathbf{i}(24-35) - \mathbf{j}(16-15) + \mathbf{k}(14-9) = \langle -11, -1, 5 \rangle$
For simplicity, we can use $\langle 11, 1, -5 \rangle$ as the normal vector $\vec{n}_Q$.

The plane $P$ contains the line $\frac{x-3}{9}= \frac{y+4}{-1}= \frac{z-7}{-5}$ (direction $\vec{v} = \langle 9, -1, -5 \rangle$) and is perpendicular to the plane with normal $\vec{n}_Q$. Therefore, the normal vector $\vec{n}_P$ is the cross product of $\vec{n}_Q$ and $\vec{v}$:
$\vec{n}_P = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 11 & 1 & -5 \\ 9 & -1 & -5 \end{vmatrix} = \mathbf{i}(-5-5) - \mathbf{j}(-55+45) + \mathbf{k}(-11-9) = \langle -10, 10, -20 \rangle$
Scaling this vector, we find $\vec{n}_P = \langle 1, -1, 2 \rangle$. Given that the plane passes through the point $(3, -4, 7)$ from the line it contains, its equation is $1(x-3) - 1(y+4) + 2(z-7) = 0$, which simplifies to $x - y + 2z - 21 = 0$.