If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_2: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$ and $\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}$ is $\sqrt{2}$, is

Let the direction cosines of two lines satisfy the equations : $4 l+m-n=0$ and $2 m n+10 n l+3 l m=0$.

Then the cosine of the acute angle between these lines is :

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{\mathrm{z}-2}{3}$. The coordinates of A and $B$ are $(1,6,3)$ and $(4,9, \alpha)$ respectively and $C$ is at a distance of 10 units from $B$. The area (in sq. units) of $\triangle A B C$ is :

Let L be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ on L , whose distance from the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line $L$ is 7 . Then $\sum\limits_{(a, b, c) \in S}(a+b+c)$ is equal to :

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$, is equal to

If the image of the point $\mathrm{P}(1,2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-\mathrm{z}}{2}$ is $\mathrm{Q}(5, b, \mathrm{c})$, then $a^2+b^2+c^2$ is equal to

Let the line L pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :

Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $\left|\overrightarrow{CD}\right|^2$ is equal to:

Let the foot of perpendicular from the point $(\lambda, 2,3)$ on the line $\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$ be the point ( $1, \mu, 2$ ). Then the distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ and $\frac{x-\lambda}{2}=\frac{y-\mu}{3}=\frac{z+5}{6}$ is equal to :

The shortest distance between the lines $\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}$ and $\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}$ is:

Let the image of the point $\mathrm{P}(1,6, a)$ in the line $\mathrm{L}: \frac{x}{1}=\frac{y-1}{2}=\frac{z-a+1}{b}, b>0$, be $\left(\frac{a}{3}, 0, a+c\right)$. If $\mathrm{S}(\alpha, \beta, \gamma), \alpha>0$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2 \sqrt{14}$, then $\alpha+\beta+\gamma$ is equal to:

Let a line L be perpendicular to both the lines $\mathrm{L}_1: \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\mathrm{L}_2: \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$.

If $\theta$ is the acute angle between the lines L and $\mathrm{L}_3: \frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2}$, then $\tan \theta$ is equal to:

Let a triangle PQR be such that P and Q lie on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ and are at a distance of 6 units from $R(1,2,3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta P Q R$, then $\alpha+\beta+\gamma$ is equal to :

If the distance of the point $(a, 2,5)$ from the image of the point $(1,2,7)$ in the line $\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{2}$ is 4 , then the sum of all possible values of $a$ is equal to :

The square of the distance of the point $\mathrm{P}(5,6,7)$ from the line $\frac{x-2}{2}=\frac{y-5}{3}=\frac{z-2}{4}$ is equal to:

$\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(a \hat{i}-\hat{j}), a \neq 0$ and $\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+a \hat{k})$ from the origin is :

The shortest distance between the lines

$ \vec{r}=\left(\frac{1}{3} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\frac{8}{3} \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) $

and $\vec{r}=\left(-\frac{2}{3} \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{k}}\right)+\mu(\hat{\mathrm{j}}-\hat{\mathrm{k}}), \lambda, \mu \in \mathbb{R}$, is:

If $\left(2 \alpha+1, \alpha^2-3 \alpha, \frac{\alpha-1}{2}\right)$ is the image of $(\alpha, 2 \alpha, 1)$ in the line $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1}$, then the possible value(s) of $\alpha$ is (are)

A line with direction ratios $1,-1,2$ intersects the lines $\frac{x}{2}=\frac{y}{3}=\frac{z+1}{3}$ and $\frac{x+1}{-1}=\frac{y-2}{1}=\frac{z}{4}$ at the points P and Q , respectively. If the length of the line segment PQ is $\alpha$, then $225 \alpha^2$ is equal to:

The square of the distance of the point $(-2,-8,6)$ from the line $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}$ along the line $\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2}$ is equal to:

Let the point A be the foot of perpendicular drawn from the point P$(a, b, 0)$ on the line

$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}.$

If the midpoint of the line segment PA is $\left(0, \frac{3}{4}, -\frac{1}{4}\right),$ then the value of $a^2 + b^2 + \alpha^2$ is equal to :

If the point of intersection of the lines $ \frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} $ and $ \frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} $ lies on xy-plane, then the value of $a+b$ is:

Let a line L passing through the point (1, 1, 1) be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of perpendicular from the origin on the line L, then the value of $34(a + b + c)$ is :

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$
and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the
points $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

If the equation of the line passing through the point $ \left( 0, -\frac{1}{2}, 0 \right) $ and perpendicular to the lines $ \vec{r} = \lambda \left( \hat{i} + a\hat{j} + b\hat{k} \right) $ and $ \vec{r} = \left( \hat{i} - \hat{j} - 6\hat{k} \right) + \mu \left( -b \hat{i} + a\hat{j} + 5\hat{k} \right) $ is $ \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} $, then $ a+b+c+d $ is equal to :

Consider the lines L₁: x - 1 = y - 2 = z and L₂: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L₁ and L₂ be Q and R respectively. If the area of the triangle PQR is A, then 4A² is equal to :

Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$. Then, which of the following points lies on the line $L$ ?

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is

Let A be the point of intersection of the lines $\mathrm{L}_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $\mathrm{L}_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let B and C be the points on the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ respectively such that $A B=A C=\sqrt{15}$. Then the square of the area of the triangle $A B C$ is :

Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ is $\frac{1}{\sqrt{6}}$, be $\mathrm{a}, \mathrm{b},(\mathrm{a}<\mathrm{b})$. Then the length of the latus rectum of the ellipse $\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$ is :

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2 \sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{O A} \cdot \overrightarrow{O B}$ is equal to

Let a line passing through the point $(4,1,0)$ intersect the line $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $\mathrm{L}_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c\end{array}\right|$ is equal to

Line $L_1$ passes through the point $(1,2,3)$ and is parallel to $z$-axis. Line $L_2$ passes through the point $(\lambda, 5,6)$ and is parallel to $y$-axis. Let for $\lambda=\lambda_1, \lambda_2, \lambda_2<\lambda_1$, the shortest distance between the two lines be 3 . Then the square of the distance of the point $\left(\lambda_1, \lambda_2, 7\right)$ from the line $L_1$ is

Let the vertices Q and R of the triangle PQR lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}, \mathrm{QR}=5$ and the coordinates of the point $P$ be $(0,2,3)$. If the area of the triangle $P Q R$ is $\frac{m}{n}$ then :

Let $A B C D$ be a tetrahedron such that the edges $A B, A C$ and $A D$ are mutually perpendicular. Let the areas of the triangles $\mathrm{ABC}, \mathrm{ACD}$ and ADB be 5,6 and 7 square units respectively. Then the area (in square units) of the $\triangle B C D$ is equal to :

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} $ and $ \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4} $. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

Let P be the foot of the perpendicular from the point $(1,2,2)$ on the line $\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$.
Let the line $\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ})^2$ is equal to :

Let $\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ be two lines.

Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $\mathrm{L}_1$, then $|5 \alpha-11 \beta-8 \gamma|$ equals :

The square of the distance of the point $ \left( \frac{15}{7}, \frac{32}{7}, 7 \right) $ from the line $ \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} $ in the direction of the vector $ \hat{i} + 4\hat{j} + 7\hat{k} $ is:

Let $\mathrm{A}(x, y, z)$ be a point in $x y$-plane, which is equidistant from three points $(0,3,2),(2,0,3)$ and $(0,0,1)$.

Let $\mathrm{B}=(1,4,-1)$ and $\mathrm{C}=(2,0,-2)$. Then among the statements

(S1) : $\triangle \mathrm{ABC}$ is an isosceles right angled triangle, and

(S2) : the area of $\triangle \mathrm{ABC}$ is $\frac{9 \sqrt{2}}{2}$,

If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

Let in a $\triangle A B C$, the length of the side $A C$ be 6 , the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle A B C$ is:

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is

If the square of the shortest distance between the lines $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}$ is $\frac{m}{n}$, where $m$, $n$ are coprime numbers, then $m+n$ is equal to :

The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is :

Let P be the foot of the perpendicular from the point $\mathrm{Q}(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $(3,-2,1)$, is

Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3 \hat{i}+2 \hat{j}+2 \hat{k}$. If the distance of the point Q from the point $\mathrm{R}(1,3,3)$ is 5 , then the square of the area of $\triangle P Q R$ is equal to :

The perpendicular distance, of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point $\mathrm{P}(2,-10,1)$, is :

Let $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $\mathrm{L}_1$ and $\mathrm{L}_2$ ?

Consider the line $\mathrm{L}$ passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $\mathrm{L}$ along the line $\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$ is equal to

The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is:

Let the line $\mathrm{L}$ intersect the lines $x-2=-y=z-1,2(x+1)=2(y-1)=z+1$ and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $\mathrm{L}$ ?

If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :

Let $P(x, y, z)$ be a point in the first octant, whose projection in the $x y$-plane is the point $Q$. Let $O P=\gamma$; the angle between $O Q$ and the positive $x$-axis be $\theta$; and the angle between $O P$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is

If the shortest distance between the lines

$\begin{array}{ll} L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\ L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, & \mu \in \mathbb{R} \end{array}$

is $\frac{m}{\sqrt{n}}$, where $\operatorname{gcd}(m, n)=1$, then the value of $m+n$ equals

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(3,-3,1)$ in the line $\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$ and $\mathrm{R}$ be the point $(2,5,-1)$. If the area of the triangle $\mathrm{PQR}$ is $\lambda$ and $\lambda^2=14 \mathrm{~K}$, then $\mathrm{K}$ is equal to :

If $A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$ and $D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$ are the vertices of a quadrilateral $A B C D$, then its area is

The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ is

Let $(\alpha, \beta, \gamma)$ be the image of the point $(8,5,7)$ in the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$. Then $\alpha+\beta+\gamma$ is equal to :

If the line $\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$, then $4 \lambda+9 \mu$ is equal to :

Let $\mathrm{d}$ be the distance of the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $(7,8,9)$. Then $\mathrm{d}^2+6$ is equal to :

Let $\mathrm{P}$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then, the shortest distance of $\mathrm{P}$ from the line $4 x=2 y=z$ is

Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at a distance of 9 units from the point $P$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Let $(\alpha, \beta, \gamma)$ be the mirror image of the point $(2,3,5)$ in the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$. Then, $2 \alpha+3 \beta+4 \gamma$ is equal to

The shortest distance, between lines $L_1$ and $L_2$, where $L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_2$ is the line, passing through the points $\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$, is

Let $L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}$,

$L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}$

be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then, the point which lies on $L_3$ is

Let $(\alpha, \beta, \gamma)$ be the foot of perpendicular from the point $(1,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. Then $19(\alpha+\beta+\gamma)$ is equal to :

Let $A(2,3,5)$ and $C(-3,4,-2)$ be opposite vertices of a parallelogram $A B C D$. If the diagonal $\overrightarrow{\mathrm{BD}}=\hat{i}+2 \hat{j}+3 \hat{k}$, then the area of the parallelogram is equal to :

Let $\mathrm{P}(3,2,3), \mathrm{Q}(4,6,2)$ and $\mathrm{R}(7,3,2)$ be the vertices of $\triangle \mathrm{PQR}$. Then, the angle $\angle \mathrm{QPR}$ is

Let $O$ be the origin and the position vectors of $A$ and $B$ be $2 \hat{i}+2 \hat{j}+\hat{k}$ and $2 \hat{i}+4 \hat{j}+4 \hat{k}$ respectively. If the internal bisector of $\angle \mathrm{AOB}$ meets the line $\mathrm{AB}$ at $\mathrm{C}$, then the length of $O C$ is

Let $P Q R$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the mid point of $\mathrm{PQ}$. The distance of the centroid of $\triangle \mathrm{PQR}$ from the point of intersection of the lines $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$ is

Let the image of the point $(1,0,7)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ be the point $(\alpha, \beta, \gamma)$. Then which one of the following points lies on the line passing through $(\alpha, \beta, \gamma)$ and making angles $\frac{2 \pi}{3}$ and $\frac{3 \pi}{4}$ with $y$-axis and $z$-axis respectively and an acute angle with $x$-axis ?

The line, that is coplanar to the line $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$, is :

The plane, passing through the points $(0,-1,2)$ and $(-1,2,1)$ and parallel to the line passing through $(5,1,-7)$ and $(1,-1,-1)$, also passes through the point :

Let $\mathrm{N}$ be the foot of perpendicular from the point $\mathrm{P}(1,-2,3)$ on the line passing through the points $(4,5,8)$ and $(1,-7,5)$. Then the distance of $N$ from the plane $2 x-2 y+z+5=0$ is :

Let the equation of plane passing through the line of intersection of the planes $x+2 y+a z=2$ and $x-y+z=3$ be $5 x-11 y+b z=6 a-1$. For $c \in \mathbb{Z}$, if the distance of this plane from the point $(a,-c, c)$ is $\frac{2}{\sqrt{a}}$, then $\frac{a+b}{c}$ is equal to :

The distance of the point $(-1,2,3)$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$ parallel to the line of the shortest distance between the lines $\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$ and $\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$ is :

Let the lines $l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ and $l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$ be coplanar. If the point $\mathrm{P}(a, b, c)$ on $l_{1}$ is nearest to the point $\mathrm{Q}(-4,-3,2)$, then $|a|+|b|+|c|$ is equal to

Let the plane P: $4 x-y+z=10$ be rotated by an angle $\frac{\pi}{2}$ about its line of intersection with the plane $x+y-z=4$. If $\alpha$ is the distance of the point $(2,3,-4)$ from the new position of the plane $\mathrm{P}$, then $35 \alpha$ is equal to :

Let the line passing through the points $\mathrm{P}(2,-1,2)$ and $\mathrm{Q}(5,3,4)$ meet the plane $x-y+z=4$ at the point $\mathrm{R}$. Then the distance of the point $\mathrm{R}$ from the plane $x+2 y+3 z+2=0$ measured parallel to the line $\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$ is equal to :

Let P be the plane passing through the points $(5,3,0),(13,3,-2)$ and $(1,6,2)$. For $\alpha \in \mathbb{N}$, if the distances of the points $\mathrm{A}(3,4, \alpha)$ and $\mathrm{B}(2, \alpha, a)$ from the plane P are 2 and 3 respectively, then the positive value of a is :

Let $(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{P}(2,3,5)$ in the plane $2 x+y-3 z=6$. Then $\alpha+\beta+\gamma$ is equal to :

If equation of the plane that contains the point $(-2,3,5)$ and is perpendicular to each of the planes $2 x+4 y+5 z=8$ and $3 x-2 y+3 z=5$ is $\alpha x+\beta y+\gamma z+97=0$ then $\alpha+\beta+\gamma=$

Let the image of the point $\mathrm{P}(1,2,6)$ in the plane passing through the points $\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)$ and $\mathrm{C}(0,5,1)$ be $\mathrm{Q}(\alpha, \beta, \gamma)$. Then $\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)$ is equal to :

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $\mathrm{A}$ and $\mathrm{B}$ respectively. Then the distance of the mid-point of the line segment $\mathrm{AB}$ from the plane $2 x-2 y+z=14$ is :

The shortest distance between the lines ${{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}$ and ${{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}$ is :

Let two vertices of a triangle ABC be (2, 4, 6) and (0, $-$2, $-$5), and its centroid be (2, 1, $-$1). If the image of the third vertex in the plane $x+2y+4z=11$ is $(\alpha,\beta,\gamma)$, then $\alpha\beta+\beta\gamma+\gamma\alpha$ is equal to :

Let P be the point of intersection of the line ${{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}$ and the plane $x+y+z=2$. If the distance of the point P from the plane $3x - 4y + 12z = 32$ is q, then q and 2q are the roots of the equation :

For $\mathrm{a}, \mathrm{b} \in \mathbb{Z}$ and $|\mathrm{a}-\mathrm{b}| \leq 10$, let the angle between the plane $\mathrm{P}: \mathrm{ax}+y-\mathrm{z}=\mathrm{b}$ and the line $l: x-1=\mathrm{a}-y=z+1$ be $\cos ^{-1}\left(\frac{1}{3}\right)$. If the distance of the point $(6,-6,4)$ from the plane P is $3 \sqrt{6}$, then $a^{4}+b^{2}$ is equal to :

Let $\mathrm{P}$ be the plane passing through the line

$\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and the point $(2,4,-3)$.

If the image of the point $(-1,3,4)$ in the plane P

is $(\alpha, \beta, \gamma)$ then $\alpha+\beta+\gamma$ is equal to :

The shortest distance between the lines $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ and $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$ is :

If the equation of the plane containing the line

$x+2 y+3 z-4=0=2 x+y-z+5$ and perpendicular to the plane

$\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$

is $a x+b y+c z=4$, then $(a-b+c)$ is equal to :

A plane P contains the line of intersection of the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$. If $\mathrm{P}$ passes through the point $(0,2,-2)$, then the square of distance of the point $(12,12,18)$ from the plane $\mathrm{P}$ is :

Let the line $\mathrm{L}$ pass through the point $(0,1,2)$, intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2 x+y-3 z=4$. Then the distance of the point $\mathrm{P}(1,-9,2)$ from the line $\mathrm{L}$ is :

If the equation of the plane passing through the line of intersection of the planes $2 x-y+z=3,4 x-3 y+5 z+9=0$ and parallel to the line $\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$ is $a x+b y+c z+6=0$, then $a+b+c$ is equal to :

One vertex of a rectangular parallelopiped is at the origin $\mathrm{O}$ and the lengths of its edges along $x, y$ and $z$ axes are $3,4$ and $5$ units respectively. Let $\mathrm{P}$ be the vertex $(3,4,5)$. Then the shortest distance between the diagonal OP and an edge parallel to $\mathrm{z}$ axis, not passing through $\mathrm{O}$ or $\mathrm{P}$ is :

Let the plane P pass through the intersection of the planes $2x+3y-z=2$ and $x+2y+3z=6$, and be perpendicular to the plane $2x+y-z+1=0$. If d is the distance of P from the point ($-$7, 1, 1), then $\mathrm{d^{2}}$ is equal to :

The shortest distance between the lines

${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$ and

${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$ is :

Let the image of the point $P(2,-1,3)$ in the plane $x+2 y-z=0$ be $Q$.

Then the distance of the plane $3 x+2 y+z+29=0$ from the point $Q$ is :

Let the shortest distance between the lines

$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$ and

$L_{1}: x+1=y-1=4-z$ be $2 \sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$,

then which of the following is NOT possible?

The line $l_1$ passes through the point (2, 6, 2) and is perpendicular to the plane $2x+y-2z=10$. Then the shortest distance between the line $l_1$ and the line $\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$ is :

The plane $2x-y+z=4$ intersects the line segment joining the points A ($a,-2,4)$ and B ($2,b,-3)$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $\sqrt5$. If $ab < 0$ and P is the point $(a-b,b,2b-a)$ then CP$^2$ is equal to :

If the lines ${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1}$ and ${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is :

The shortest distance between the lines ${{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$ and ${{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$ is :

The foot of perpendicular of the point (2, 0, 5) on the line ${{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}}$ is ($\alpha,\beta,\gamma$). Then, which of the following is NOT correct?

The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is :

The distance of the point P(4, 6, $-$2) from the line passing through the point ($-$3, 2, 3) and parallel to a line with direction ratios 3, 3, $-$1 is equal to :

Consider the lines $L_1$ and $L_2$ given by

${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$

${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$.

A line $L_3$ having direction ratios 1, $-$1, $-$2, intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively. Then the length of line segment $PQ$ is

If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is ($\alpha,\beta,\gamma$), then $\alpha+\beta+\gamma$ is equal to :

Let the plane containing the line of intersection of the planes

P₁ : $x+(\lambda+4)y+z=1$ and

P₂ : $2x+y+z=2$

pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of

the point (2$\lambda,\lambda,-\lambda$) from the plane P₂ is :

The distance of the point (7, $-$3, $-$4) from the plane passing through the points (2, $-$3, 1), ($-$1, 1, $-$2) and (3, $-$4, 2) is :

The distance of the point ($-1,9,-16$) from the plane

$2x+3y-z=5$ measured parallel to the line

${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$ is :

Let $Q$ be the foot of perpendicular drawn from the point $P(1,2,3)$ to the plane $x+2 y+z=14$. If $R$ is a point on the plane such that $\angle P R Q=60^{\circ}$, then the area of $\triangle P Q R$ is equal to :

If $(2,3,9),(5,2,1),(1, \lambda, 8)$ and $(\lambda, 2,3)$ are coplanar, then the product of all possible values of $\lambda$ is:

If the foot of the perpendicular from the point $\mathrm{A}(-1,4,3)$ on the plane $\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4$, is $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $3,-1,-4$, is equal to :

Let the lines

$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ and

$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$ be coplanar

and $\mathrm{P}$ be the plane containing these two lines.

Then which of the following points does NOT lie on P?

A plane P is parallel to two lines whose direction ratios are $-2,1,-3$ and $-1,2,-2$ and it contains the point $(2,2,-2)$. Let P intersect the co-ordinate axes at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ making the intercepts $\alpha, \beta, \gamma$. If $\mathrm{V}$ is the volume of the tetrahedron $\mathrm{OABC}$, where $\mathrm{O}$ is the origin, and $\mathrm{p}=\alpha+\beta+\gamma$, then the ordered pair $(\mathrm{V}, \mathrm{p})$ is equal to :

The foot of the perpendicular from a point on the circle $x^{2}+y^{2}=1, z=0$ to the plane $2 x+3 y+z=6$ lies on which one of the following curves?

If the length of the perpendicular drawn from the point $P(a, 4,2)$, a $>0$ on the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$ is $2 \sqrt{6}$ units and $Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ is the image of the point P in this line, then $\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}$ is equal to :

If the line of intersection of the planes $a x+b y=3$ and $a x+b y+c z=0$, a $>0$ makes an angle $30^{\circ}$ with the plane $y-z+2=0$, then the direction cosines of the line are :

If the plane $P$ passes through the intersection of two mutually perpendicular planes $2 x+k y-5 z=1$ and $3 k x-k y+z=5, k<3$ and intercepts a unit length on positive $x$-axis, then the intercept made by the plane $P$ on the $y$-axis is :

A vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i}+\hat{j}$ and the plane determined by the vectors $\hat{i}-\hat{j}, \hat{i}+\hat{k}$. The obtuse angle between $\vec{a}$ and the vector $\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$ is :

The length of the perpendicular from the point $(1,-2,5)$ on the line passing through $(1,2,4)$ and parallel to the line $x+y-z=0=x-2 y+3 z-5$ is :

A plane $E$ is perpendicular to the two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, and passes through the point $P(1,-1,1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3 \sqrt{2}$, then $(P Q)^{2}$ is equal to :

The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=z$ and $\frac{7-x}{2}=y-2=z-6$ is :

Let $\mathrm{P}$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$. If $\mathrm{d}$ is the distance of $\mathrm{P}$ from the point $(2,-5,11)$, then $\mathrm{d}^{2}$ is equal to :

The distance of the point (3, 2, $-$1) from the plane $3x - y + 4z + 1 = 0$ along the line ${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$ is equal to :

Let ${{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}}$ lie on the plane $px - qy + z = 5$, for some p, q $\in$ R. The shortest distance of the plane from the origin is :

Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line $\overrightarrow r = - \widehat k + \lambda \left( {\widehat i + \widehat j + 2\widehat k} \right),\,\lambda \in R$. Then, which of the following points lies on T?

If the mirror image of the point (2, 4, 7) in the plane 3x $-$ y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to :

Let the plane ax + by + cz = d pass through (2, 3, $-$5) and is perpendicular to the planes
2x + y $-$ 5z = 10 and 3x + 5y $-$ 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :

If two distinct point Q, R lie on the line of intersection of the planes $ - x + 2y - z = 0$ and $3x - 5y + 2z = 0$ and $PQ = PR = \sqrt {18} $ where the point P is (1, $-$2, 3), then the area of the triangle PQR is equal to :

The acute angle between the planes P₁ and P₂, when P₁ and P₂ are the planes passing through the intersection of the planes $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the points (2, 1, 3) and (0, 1, 2), respectively, is :

Let the plane $P:\overrightarrow r \,.\,\overrightarrow a = d$ contain the line of intersection of two planes $\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 6$ and $\overrightarrow r \,.\,\left( { - 6\widehat i + 5\widehat j - \widehat k} \right) = 7$. If the plane P passes through the point $\left( {2,3,{1 \over 2}} \right)$, then the value of ${{|13\overrightarrow a {|^2}} \over {{d^2}}}$ is equal to :

Let the foot of the perpendicular from the point (1, 2, 4) on the line ${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$ be P. Then the distance of P from the plane $3x + 4y + 12z + 23 = 0$ is :

The shortest distance between the lines

${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$ and ${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$, is :

If two straight lines whose direction cosines are given by the relations $l + m - n = 0$, $3{l^2} + {m^2} + cnl = 0$ are parallel, then the positive value of c is :

If the plane $2x + y - 5z = 0$ is rotated about its line of intersection with the plane $3x - y + 4z - 7 = 0$ by an angle of ${\pi \over 2}$, then the plane after the rotation passes through the point :

If the lines $\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right)$ and $\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right)$ are co-planar, then the distance of the plane containing these two lines from the point ($\alpha$, 0, 0) is :

Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$, $\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$ and $\overrightarrow c = \widehat i - \widehat j + \widehat k$ be three given vectors. Let $\overrightarrow v $ be a vector in the plane of $\overrightarrow a $ and $\overrightarrow b $ whose projection on $\overrightarrow c $ is ${2 \over {\sqrt 3 }}$. If $\overrightarrow v \,.\,\widehat j = 7$, then $\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)$ is equal to :

If the two lines ${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$ and ${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$ are perpendicular, then an angle between the lines l₂ and ${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$ is :

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $-$ 3y + 5z = 8. If the mirror image of the point $\left( {2, - {1 \over 2},2} \right)$ in the rotated plane is B(a, b, c), then :

Let p be the plane passing through the intersection of the planes $\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 5$ and $\overrightarrow r \,.\,\left( {2\widehat i - \widehat j + \widehat k} \right) = 3$, and the point (2, 1, $-$2). Let the position vectors of the points X and Y be $\widehat i - 2\widehat j + 4\widehat k$ and $5\widehat i - \widehat j + 2\widehat k$ respectively. Then the points :

Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $-$1, $-$1), parallel to the line PQ meets the plane S at R, then QR² is equal to :

If the shortest distance between the lines ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }$ and ${{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}$ is ${1 \over {\sqrt 3 }}$, then the sum of all possible value of $\lambda$ is :

Let the points on the plane P be equidistant from the points ($-$4, 2, 1) and (2, $-$2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :