Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The angle between two planes is determined by the angle between their normal vectors, which relates to the dot product of the normals through the formula $\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$. Given the sine of the angle $\sin\theta = \frac{2\sqrt{6}}{5}$, we identify $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{24}{25}} = \frac{1}{5}$. The normal vectors are $\vec{n}_1 = 3\hat{i} - 5\hat{j} + \hat{k}$ and $\vec{n}_2 = \lambda\hat{i} + \hat{j} - 3\hat{k}$, with respective magnitudes $|\vec{n}_1| = \sqrt{3^2 + (-5)^2 + 1^2} = \sqrt{35}$ and $|\vec{n}_2| = \sqrt{\lambda^2 + 1^2 + (-3)^2} = \sqrt{\lambda^2 + 10}$.

Substituting these values into the cosine equation yields $\frac{1}{5} = \frac{|3\lambda - 8|}{\sqrt{35}\sqrt{\lambda^2 + 10}}$. Squaring both sides produces $\frac{1}{25} = \frac{(3\lambda - 8)^2}{35(\lambda^2 + 10)}$, which simplifies to $7(\lambda^2 + 10) = 5(9\lambda^2 - 48\lambda + 64)$. Expanding this leads to the quadratic equation $19\lambda^2 - 120\lambda + 125 = 0$. Factoring this as $(19\lambda - 25)(\lambda - 5) = 0$ provides the two values $\lambda = \frac{25}{19}$ and $\lambda = 5$. Since $\lambda_1 < \lambda_2$, we set $\lambda_1 = \frac{25}{19}$ and $\lambda_2 = 5$.

The point $(38\lambda_1, 10\lambda_2, 2)$ becomes $(38 \cdot \frac{25}{19}, 10 \cdot 5, 2)$, which simplifies to $(50, 50, 2)$. The perpendicular distance from a point $(x_0, y_0, z_0)$ to the plane $ax + by + cz = d$ is given by $\frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$. For the plane $3x - 5y + z = 7$, the distance is $\frac{|3(50) - 5(50) + 1(2) - 7|}{\sqrt{3^2 + (-5)^2 + 1^2}} = \frac{|150 - 250 + 2 - 7|}{\sqrt{35}} = \frac{|-105|}{\sqrt{35}} = \frac{105}{\sqrt{35}}$. Squaring this distance results in $\frac{105^2}{35} = \frac{11025}{35} = 315$.

The shortest distance between two skew lines is determined by the projection of the vector connecting any point on the first line to any point on the second line onto the common perpendicular vector. Converting the provided lines $x+1=2y=-12z$ and $x=y+2=6z-6$ into symmetric form produces the expressions:
$\frac{x+1}{1}=\frac{y}{1/2}=\frac{z}{-1/12} \quad \text{and} \quad \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{1/6}$
This identification yields point $A=(-1, 0, 0)$ with direction $\vec{p}=(1, 1/2, -1/12)$ and point $B=(0, -2, 1)$ with direction $\vec{q}=(1, 1, 1/6)$. The common perpendicular vector $\vec{n} = \vec{p} \times \vec{q}$ simplifies to $\vec{n} = (2, -3, 6)$, and using the vector $\vec{BA} = (-1, 2, -1)$ that connects the two lines allows for the distance calculation via the projection formula:
$\text{Distance} = \frac{|\vec{BA} \cdot \vec{n}|}{|\vec{n}|} = \frac{|(-1)(2) + (2)(-3) + (-1)(6)|}{\sqrt{2^{2+}(-3)^{2+}6^2}} = \frac{|-14|}{7} = 2$

The shortest distance between two skew lines is determined by the scalar projection of the vector connecting a point on each line onto the common perpendicular vector shared by both lines. The lines $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ and $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$ possess respective points $\vec{a}=(4,-2,-3)$ and $\vec{b}=(1,3,4)$ with corresponding direction vectors $\vec{n}_1=(4,5,3)$ and $\vec{n}_2=(3,4,2)$.

The common perpendicular vector is obtained via the cross product $\vec{n}_1 \times \vec{n}_2 = (-2,1,1)$, which has a magnitude of $\sqrt{6}$. With the vector connecting the points given by $\vec{a}-\vec{b}=(3,-5,-7)$, the shortest distance $d$ is calculated by projecting this connector onto the perpendicular normal vector using the formula:

$d = \frac{|(\vec{a}-\vec{b}) \cdot (\vec{n}_1 \times \vec{n}_2)|}{|\vec{n}_1 \times \vec{n}_2|} = \frac{|(3,-5,-7) \cdot (-2,1,1)|}{\sqrt{6}}$

Evaluating this expression results in $d = \frac{|-6-5-7|}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3\sqrt{6}$.

For a line to be contained within a plane, two geometric conditions must be satisfied simultaneously: every point on the line must lie on the plane, and the direction vector of the line must be orthogonal to the normal vector of the plane. The given line equation $x+10 = \frac{8-y}{2} = z$ can be expressed in symmetric form as $\frac{x+10}{1} = \frac{y-8}{-2} = \frac{z}{1}$, identifying a point $(-10, 8, 0)$ and a direction vector $\langle 1, -2, 1 \rangle$. Substituting the point into the plane equation $ax + by + 3z = 2(a+b)$ yields $-10a + 8b = 2a + 2b$, which simplifies to the relationship $b=2a$. Furthermore, requiring the normal vector $\langle a, b, 3 \rangle$ to be perpendicular to the line's direction vector $\langle 1, -2, 1 \rangle$ implies that their dot product is zero, resulting in the equation $a - 2b + 3 = 0$. Substituting $b=2a$ into this condition leads to $a - 2(2a) + 3 = 0$, which solves to $a=1$ and $b=2$.

With the plane equation determined as $x + 2y + 3z - 6 = 0$, the perpendicular distance $c$ from the point $(1, 27, 7)$ to the plane is calculated using the distance formula:
$c = \frac{|1(1) + 2(27) + 3(7) - 6|}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{|1 + 54 + 21 - 6|}{\sqrt{14}} = \frac{70}{\sqrt{14}} = 5\sqrt{14}$
Substituting these values into the final expression $a^2 + b^2 + c^2$ gives $1^2 + 2^2 + (5\sqrt{14})^2$, which is $1 + 4 + 350$, resulting in 355.

To determine the foot of a perpendicular from a point to a line, we first establish the direction of the line by finding the cross product of the normal vectors of the two given parallel planes, $\langle 1, 2, 1 \rangle$ and $\langle 0, 3, -1 \rangle$, which results in the direction vector $\langle -5, 1, 3 \rangle$. With the line passing through the point $(3, 2, 1)$, any general point $M$ on the line can be expressed as $(3 - 5\lambda, 2 + \lambda, 1 + 3\lambda)$. Connecting the point $P(1, 9, 7)$ to this general point $M$ yields the vector $\vec{PM} = \langle 2 - 5\lambda, \lambda - 7, 3\lambda - 6 \rangle$.

For $M$ to be the foot of the perpendicular, the vector $\vec{PM}$ must be orthogonal to the line's direction vector, meaning their dot product must be zero:
$-5(2 - 5\lambda) + 1(\lambda - 7) + 3(3\lambda - 6) = 0$
Expanding this equation results in $-10 + 25\lambda + \lambda - 7 + 9\lambda - 18 = 0$, which simplifies to $35\lambda - 35 = 0$, confirming that $\lambda = 1$. Substituting this parameter value back into the coordinates for $M$ gives $(-2, 3, 4)$, and calculating the sum of these coordinates yields $-2 + 3 + 4 = 5$.

The shortest distance between two skew lines is determined by projecting the vector connecting a point on each line onto the common perpendicular, a vector found by taking the cross product of the lines' direction vectors. From the line equations, the lines pass through the points $(5, 2, 4)$ and $(-3, -5, 1)$, resulting in the displacement vector $(8, 7, 3)$, with respective direction vectors $(1, 2, -3)$ and $(1, 4, -5)$.

The calculation requires dividing the scalar triple product of these vectors by the magnitude of the cross product of the two direction vectors. The scalar triple product corresponds to the determinant

, which evaluates to 36. The cross product of the direction vectors results in the vector $(2, 2, 2)$, which has a magnitude of $\sqrt{2^2 + 2^2 + 2^2}$, equivalent to $2\sqrt{3}$. Dividing 36 by $2\sqrt{3}$ results in $6\sqrt{3}$.

The shortest distance between two skew lines is found by projecting the vector connecting a point on each line onto their common perpendicular, which is the cross product of their respective direction vectors. The diagonal $OP$ originates at $(0,0,0)$ and ends at $(3,4,5)$, providing a direction vector of $\mathbf{u} = (3,4,5)$. An edge parallel to the $z$-axis that avoids the origin and the point $P$ can be represented by the line passing through $(3,0,0)$ with a direction vector $\mathbf{v} = (0,0,1)$. Calculating the cross product $\mathbf{u} \times \mathbf{v} = (4, -3, 0)$ yields a vector with a magnitude of $\sqrt{4^2 + (-3)^2} = 5$. The numerator of the distance formula is given by the absolute value of the scalar triple product, calculated as the determinant

. Dividing this result by the magnitude of the cross product gives the shortest distance as $\frac{12}{5}$.

To find the intersection point $C$, first establish the parametric form for the line passing through $A(-3, -6, 1)$ and $B(2, 4, -3)$. The direction vector of this line is $(2 - (-3), 4 - (-6), -3 - 1)$, which is $(5, 10, -4)$. Using point $B(2, 4, -3)$, any point on this line can be written as $(5\lambda + 2, 10\lambda + 4, -4\lambda - 3)$. Substituting these coordinates into the plane equation $8x + y + 2z = 0$ gives $8(5\lambda + 2) + (10\lambda + 4) + 2(-4\lambda - 3) = 0$, which simplifies to $50\lambda + 16 + 10\lambda + 4 - 8\lambda - 6 = 0$. Solving $42\lambda = -14$ results in $\lambda = -1/3$. Plugging $\lambda = -1/3$ back into the parametric equations yields the point $C$ as $(1/3, 2/3, -5/3)$.

With point $C$ identified, consider the target line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$, which is equivalent to $\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3} = \mu$. Any point $D$ on this line has coordinates $(-\mu + 1, 2\mu - 4, 3\mu - 2)$. The vector $\vec{CD}$ represents the perpendicular from $C$ to the line, with components $(-\mu + 1 - 1/3, 2\mu - 4 - 2/3, 3\mu - 2 + 5/3)$, which simplifies to $(-\mu + 2/3, 2\mu - 14/3, 3\mu - 1/3)$. Since this vector must be perpendicular to the line's direction vector $(-1, 2, 3)$, their dot product is zero: $(-1)(-\mu + 2/3) + 2(2\mu - 14/3) + 3(3\mu - 1/3) = 0$. Expanding this gives $\mu - 2/3 + 4\mu - 28/3 + 9\mu - 1 = 0$, which simplifies to $14\mu = 11$, or $\mu = 11/14$.

Substituting $\mu = 11/14$ back into the vector components gives $( -11/14 + 2/3, 2(11/14) - 14/3, 3(11/14) - 1/3 )$. Calculating these values using a common denominator of $42$ results in $(-5/42, -130/42, 85/42)$. Scaling these by $-42/5$ to obtain integer direction ratios gives $(-1, -26, 17)$. Setting $a = -1$, $b = -26$, and $c = 17$, the expression $|a + b + c|$ becomes $|-1 - 26 + 17|$, which is $|-10|$, resulting in a final value of $10$.

The distance of any point $(x_0, y_0, z_0)$ from the plane $Ax + By + Cz + D = 0$ is calculated using the formula $\frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^{2+}B^{2+}C^2}}$. By setting this value equal for the points $(\frac{5}{2}, 1, \lambda)$ and $(-2, 0, 1)$ relative to the plane $2x + 3y - 6z + 7 = 0$, the identical denominator of $\sqrt{2^2 + 3^2 + (-6)^2} = 7$ allows us to simplify the expression to $|2(\frac{5}{2}) + 3(1) - 6\lambda + 7| = |2(-2) + 3(0) - 6(1) + 7|$. This reduces to the equation $|15 - 6\lambda| = |-3| = 3$. Solving for $\lambda$ produces two possible values, $15 - 6\lambda = 3$ or $15 - 6\lambda = -3$, resulting in $\lambda = 2$ and $\lambda = 3$. Since the condition $\lambda_1 > \lambda_2$ is specified, we set $\lambda_1 = 3$ and $\lambda_2 = 2$.

Using these values, the point becomes $(\lambda_1 - \lambda_2, \lambda_2, \lambda_1) = (3 - 2, 2, 3) = (1, 2, 3)$. The distance from a point $P$ to a line passing through point $A$ with direction vector $\vec{b}$ is determined by $\frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$. The line $\frac{x-5}{1} = \frac{y-1}{2} = \frac{z+7}{2}$ passes through $A(5, 1, -7)$ and has a direction vector $\vec{b} = (1, 2, 2)$. Determining the vector $\vec{AP} = (1-5, 2-1, 3-(-7)) = (-4, 1, 10)$, we compute the cross product $\vec{AP} \times \vec{b} = (-4, 1, 10) \times (1, 2, 2) = (-18, 18, -9)$. The magnitude of this result is $\sqrt{(-18)^2 + 18^2 + (-9)^2} = 27$, and the magnitude of the direction vector is $\sqrt{1^2 + 2^2 + 2^2} = 3$. Dividing the magnitudes gives the required distance of $\frac{27}{3} = 9$.

The plane is determined by its normal vector, which corresponds to the line segment from the origin to the foot of the perpendicular at $(2, a, 4)$. Using the point-normal form $2(x-2) + a(y-a) + 4(z-4) = 0$, the equation simplifies to $2x + ay + 4z = a^2 + 20$. The intercepts on the coordinate axes are $A = (\frac{a^{2+}20}{2}, 0, 0)$, $B = (0, \frac{a^{2+}20}{a}, 0)$, and $C = (0, 0, \frac{a^{2+}20}{4})$. Given the volume of the tetrahedron is 144 $\text{unit}^3$, we set $\frac{1}{6} \cdot (\frac{a^{2+}20}{2}) \cdot (\frac{a^{2+}20}{a}) \cdot (\frac{a^{2+}20}{4}) = 144$, which simplifies to $(a^{2+}20)^3 = 6912a$. Testing natural numbers for $a$, we find $a=2$ is the unique solution because $(4+20)^3 = 24^3 = 13824$ and $6912 \cdot 2 = 13824$. Substituting $a=2$ into the plane equation yields $2x + 2y + 4z = 24$, or $x + y + 2z = 12$. Among the given options, the point $(3, 0, 4)$ does not lie on the plane because $3 + 0 + 2(4) = 11 \neq 12$.

The orthocenter of a triangle in three-dimensional space is the point where its altitudes intersect, a property requiring that the vector connecting any vertex to the orthocenter is perpendicular to the side opposite that vertex. For the plane defined by $x+3y-2z+6=0$, the intercepts on the coordinate axes provide the vertices $A(-6, 0, 0)$, $B(0, -2, 0)$, and $C(0, 0, 3)$. With the orthocenter $H$ given as $\left(\alpha, \beta, \frac{6}{7}\right)$, one can utilize the orthogonality conditions $\vec{AH} \cdot \vec{BC} = 0$ and $\vec{CH} \cdot \vec{AB} = 0$ to solve for the unknown coordinates.

Using the determined vertices, the vector $\vec{BC}$ is $(0, 2, 3)$ and $\vec{AB}$ is $(6, -2, 0)$. The vector $\vec{AH}$ is $\left(\alpha+6, \beta, \frac{6}{7}\right)$, and the dot product $\vec{AH} \cdot \vec{BC} = 0$ leads directly to $2\beta + \frac{18}{7} = 0$, which yields $\beta = -\frac{9}{7}$. Similarly, the vector $\vec{CH}$ is $\left(\alpha, \beta, -\frac{15}{7}\right)$, and the condition $\vec{CH} \cdot \vec{AB} = 0$ simplifies to $6\alpha - 2\beta = 0$, meaning $3\alpha = \beta$. Substituting the value of $\beta$ gives $\alpha = -\frac{3}{7}$. Computing the final expression $98(\alpha+\beta)^2$ requires evaluating $98\left(-\frac{3}{7} - \frac{9}{7}\right)^2$, which simplifies to $98\left(-\frac{12}{7}\right)^2$ or $98 \cdot \frac{144}{49}$, resulting in $288$.

The distance from a point to a line is found by locating the foot of the perpendicular from the point to the line, where the vector between the point and the line is orthogonal to the line's direction vector. Given the line $L$ passes through $A(0, 1, 2)$ and intersects the line defined by $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$, any point $Q$ on the latter can be expressed as $(2\lambda+1, 3\lambda+2, 4\lambda+3)$. The direction vector of $L$ is $Q-A = (2\lambda+1, 3\lambda+1, 4\lambda+1)$. Since $L$ is parallel to the plane $2x+y-3z=4$, the direction vector must be perpendicular to the plane's normal $(2, 1, -3)$, leading to the equation $2(2\lambda+1) + 1(3\lambda+1) - 3(4\lambda+1) = 0$, which simplifies to $-5\lambda = 0$, yielding $\lambda = 0$.

At $\lambda=0$, line $L$ has the direction $(1, 1, 1)$ and passes through $(0, 1, 2)$, so its equation is $\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\mu$. A general point $R(\mu, \mu+1, \mu+2)$ lies on $L$. The vector from $P(1, -9, 2)$ to $R$ is $(\mu-1, \mu+10, \mu)$. Setting the dot product of this vector with the line's direction $(1, 1, 1)$ to zero ensures the vector is perpendicular to the line, giving $(\mu-1) + (\mu+10) + \mu = 0$, or $3\mu + 9 = 0$, implying $\mu = -3$. The distance is the magnitude of the vector $PR$ at $\mu = -3$, which is calculated as $\sqrt{(-3-1)^2 + (-3+1+9)^2 + (-3+2-2)^2} = \sqrt{(-4)^2 + 7^2 + (-3)^2} = \sqrt{16+49+9} = \sqrt{74}$.

Expressing the line in parametric form involves setting each of its ratios equal to a common variable $\lambda$, resulting in $x = 3\lambda - 3$, $y = \lambda - 2$, and $z = 1 - 2\lambda$. Substituting these expressions directly into the plane equation $x+y+z=2$ yields $3\lambda - 3 + \lambda - 2 + 1 - 2\lambda = 2$, which simplifies to $2\lambda - 4 = 2$, identifying $\lambda = 3$. Applying this parameter back to the expressions reveals the coordinates of point $P$ as $(6, 1, -5)$.

The perpendicular distance $q$ from point $P$ to the plane $3x - 4y + 12z - 32 = 0$ is found using the distance formula, which is $q = \left| \frac{3(6) - 4(1) + 12(-5) - 32}{\sqrt{3^2 + (-4)^2 + 12^2}} \right|$. Simplifying the numerator and denominator results in $q = \left| \frac{-78}{13} \right|$, so $q = 6$. Given that $q$ and $2q$ are the roots, they must be $6$ and $12$, and using these values to form a quadratic equation with a sum of $18$ and a product of $72$ leads to $x^2 - 18x + 72 = 0$.

Finding the foot of the perpendicular from a point to a line requires expressing any arbitrary point on the line in terms of a single parameter and utilizing the condition that the vector joining the given point and this arbitrary point must be orthogonal to the line's direction vector. For the line defined by $\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$, we equate the ratios to a parameter $\lambda$, allowing any point $P$ on the line to be represented as $(\alpha, \beta, \gamma) = (2\lambda-1, 5\lambda+1, -\lambda-1)$. Given the point $A(2, 0, 5)$, the vector $\vec{AP}$ is determined by subtracting the coordinates of $A$ from the coordinates of $P$, resulting in $(2\lambda-3, 5\lambda+1, -\lambda-6)$. Because this vector must be perpendicular to the line's direction vector $(2, 5, -1)$, their dot product must equal zero.

Performing the dot product $2(2\lambda-3) + 5(5\lambda+1) - 1(-\lambda-6) = 0$ simplifies to $4\lambda - 6 + 25\lambda + 5 + \lambda + 6 = 30\lambda + 5 = 0$, which yields $\lambda = -\frac{1}{6}$. Substituting this value back into the parametric expressions gives the coordinates of the foot of the perpendicular: $\alpha = 2(-\frac{1}{6}) - 1 = -\frac{4}{3}$, $\beta = 5(-\frac{1}{6}) + 1 = \frac{1}{6}$, and $\gamma = -(-\frac{1}{6}) - 1 = -\frac{5}{6}$. Evaluating the provided ratios with these coordinates confirms that $\frac{\alpha\beta}{\gamma} = \frac{(-4/3)(1/6)}{-5/6} = \frac{4}{15}$, $\frac{\alpha}{\beta} = \frac{-4/3}{1/6} = -8$, and $\frac{\gamma}{\alpha} = \frac{-5/6}{-4/3} = \frac{5}{8}$ are all accurate statements. However, the ratio $\frac{\beta}{\gamma}$ calculates to $\frac{1/6}{-5/6} = -\frac{1}{5}$, which contradicts the assertion that the ratio equals $-5$.

The shortest distance between two lines in three-dimensional space is found by projecting the vector connecting any point on the first line to any point on the second line onto the common perpendicular, which is the vector resulting from the cross product of the two lines' direction vectors. For the first line passing through $(1, 2, 3)$ and $(2, 3, 4)$, the direction vector is $\vec{p} = \hat{i} + \hat{j} + \hat{k}$ and a point on the line is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$. The second line, given by $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$, provides the direction vector $\vec{q} = 2\hat{i} - \hat{j}$ and passes through the point $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.

Calculating the cross product of the two direction vectors produces the common normal vector:
$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} = \hat{i} + 2\hat{j} - 3\hat{k}$
The magnitude of this normal vector is $\sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14}$. The vector connecting the points on the two lines is $\vec{b} - \vec{a} = (\hat{i} - \hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = -3\hat{j} - \hat{k}$. By taking the absolute value of the dot product of this displacement vector and the normal vector, then dividing by the magnitude of the normal, the shortest distance $\alpha$ is determined:
$\alpha = \left| \frac{(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right| = \left| \frac{(-3\hat{j} - \hat{k}) \cdot (\hat{i} + 2\hat{j} - 3\hat{k})}{\sqrt{14}} \right| = \frac{|-6 + 3|}{\sqrt{14}} = \frac{3}{\sqrt{14}}$
Squaring this result and multiplying by $28$ yields $28 \alpha^2 = 28 \times \frac{9}{14}$, which simplifies to $18$.

When measuring the distance of a point from a plane along a direction parallel to a given line, we construct a line passing through the point with the direction ratios of the given line and determine where it intersects the plane. The provided line $\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}$ has direction ratios $(3, -4, 12)$, which we identify by rewriting the term for $y$ as $\frac{y-2}{-4}$. We construct the line passing through $(-1, 9, -16)$ with these direction ratios, parameterized as the coordinates $(3\lambda-1, -4\lambda+9, 12\lambda-16)$.

Substituting these parametric coordinates into the plane equation $2x+3y-z=5$ leads to the equation $2(3\lambda-1)+3(-4\lambda+9)-(12\lambda-16)=5$. Expanding this results in $6\lambda-2-12\lambda+27-12\lambda+16=5$, which simplifies to $-18\lambda+41=5$, yielding $\lambda=2$. With this value, the intersection point on the plane is found to be $(5, 1, 8)$. The distance is the magnitude of the displacement between the original point $(-1, 9, -16)$ and the intersection point $(5, 1, 8)$, calculated as $\sqrt{(5 - (-1))^2 + (1 - 9)^2 + (8 - (-16))^2} = \sqrt{6^{2+}(-8)^{2+}24^2} = \sqrt{36+64+576} = \sqrt{676}$, which equals 26.