A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at $45^{\circ}$ and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$, is normal to $\vec{v}$, then
(1) $\sqrt{2}a + b + c = 1$
(2) $a + b + \sqrt{2}c = 1$
(3) $a + \sqrt{2}b + c = 1$
(4) $\sqrt{2}a - b + c = 1$

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
(1) $6 \sqrt { 3 }$
(2) $2 \sqrt { 6 }$
(3) $6 \sqrt { 2 }$
(4) $3 \sqrt { 6 }$

Consider the lines $L _ { 1 }$ and $L _ { 2 }$ given by $L _ { 1 } : \frac { x - 1 } { 2 } = \frac { y - 3 } { 1 } = \frac { z - 2 } { 2 }$ $L _ { 2 } : \frac { x - 2 } { 1 } = \frac { y - 2 } { 2 } = \frac { z - 3 } { 3 }$
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 $P Q$ is
(1) $2 \sqrt { 6 }$
(2) $3 \sqrt { 2 }$
(3) $4 \sqrt { 3 }$
(4) 4

Let the co-ordinates of one vertex of $\triangle A B C$ be $A ( 0,2 , \alpha )$ and the other two vertices lie on the line $\frac { \mathrm { x } + \alpha } { 5 } = \frac { \mathrm { y } - 1 } { 2 } = \frac { \mathrm { z } + 4 } { 3 }$. For $\alpha \in \mathbb { Z }$, if the area of $\triangle A B C$ is 21 sq. units and the line segment $B C$ has length $2 \sqrt { 21 }$ units, then $\alpha ^ { 2 }$ is equal to $\_\_\_\_$ .

If a plane passes through the points $(-1, k, 0)$, $(2, k, -1)$, $(1, 1, 2)$ and is parallel to the line $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$, then the value of $\frac{k^{2}+1}{(k-1)(k-2)}$ is
(1) $\frac{17}{5}$
(2) $\frac{5}{17}$
(3) $\frac{6}{13}$
(4) $\frac{13}{6}$

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = \sqrt{14}$, $|\vec{b}| = \sqrt{6}$ and $|\vec{a} \times \vec{b}| = \sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $\underline{\hspace{1cm}}$.

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
(1) 18
(2) 22
(3) 20
(4) 24

Let a line $L$ pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of $L$ from the point $(5, 3, 8)$ is $\alpha$, then $3\alpha^{2}$ is equal to $\_\_\_\_$

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x + y + 3z = 16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$, then $\alpha^2$ is equal to $\underline{\hspace{1cm}}$.

Let $\lambda _ { 1 } , \lambda _ { 2 }$ be the values of $\lambda$ for which the points $\left( \frac { 5 } { 2 } , 1 , \lambda \right)$ and $( - 2 , 0 , 1 )$ are at equal distance from the plane $2 x + 3 y - 6 z + 7 = 0$. If $\lambda _ { 1 } > \lambda _ { 2 }$, then the distance of the point $\left( \lambda _ { 1 } - \lambda _ { 2 } , \lambda _ { 2 } , \lambda _ { 1 } \right)$ from the line $\frac { x - 5 } { 1 } = \frac { y - 1 } { 2 } = \frac { z + 7 } { 2 }$ is $\_\_\_\_$

Let $\theta$ be the angle between the planes $P_1: \vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 9$ and $P_2: \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 15$. Let $L$ be the line that meets $P_2$ at the point $(4,-2,5)$ and makes an angle $\theta$ with the normal of $P_2$. If $\alpha$ is the angle between $L$ and $P_2$, then $\tan^2\theta \cdot \cot^2\alpha$ is equal to $\underline{\hspace{1cm}}$.

Consider a $\triangle ABC$ where $A(1,3,2)$, $B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$, then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:
(1) $\frac{37}{2\sqrt{38}}$
(2) $\frac{\sqrt{38}}{2}$
(3) $\frac{39}{2\sqrt{38}}$
(4) $\sqrt{19}$

The position vectors of the vertices $A , B$ and $C$ of a triangle are $2 \hat { i } - 3 \hat { j } + 3 \hat { k } , \quad 2 \hat { i } + 2 \hat { j } + 3 \hat { k }$ and $- \hat { i } + \hat { j } + 3 \hat { k }$ respectively. Let $l$ denotes the length of the angle bisector AD of $\angle \mathrm { BAC }$ where D is on the line segment BC , then $2 l ^ { 2 }$ equals :
(1) 49
(2) 42
(3) 50
(4) 45

Let a unit vector $\widehat { u } = x \hat { i } + y \hat { j } + z \widehat { k }$ make angles $\frac { \pi } { 2 } , \frac { \pi } { 3 }$ and $\frac { 2 \pi } { 3 }$ with the vectors $\frac { 1 } { \sqrt { 2 } } \hat { i } + \frac { 1 } { \sqrt { 2 } } \widehat { k } , \frac { 1 } { \sqrt { 2 } } \hat { j } + \frac { 1 } { \sqrt { 2 } } \widehat { k }$ and $\frac { 1 } { \sqrt { 2 } } \hat { i } + \frac { 1 } { \sqrt { 2 } } \hat { j }$ respectively. If $\vec { v } = \frac { 1 } { \sqrt { 2 } } \hat { i } + \frac { 1 } { \sqrt { 2 } } \hat { j } + \frac { 1 } { \sqrt { 2 } } \hat { k }$, then $| \hat { u } - \vec { v } | ^ { 2 }$ is equal to
(1) $\frac { 11 } { 2 }$
(2) $\frac { 5 } { 2 }$
(3) 9
(4) 7

Let the position vectors of the vertices $A , B$ and $C$ of a triangle be $2 \hat { i } + 2 \hat { j } + \hat { k } , \hat { i } + 2 \hat { j } + 2 \hat { k }$ and $2 \hat { i } + \hat { j } + 2 \hat { k }$ respectively. Let $l _ { 1 } , l _ { 2 }$ and $l _ { 3 }$ be the lengths of perpendiculars drawn from the ortho centre of the triangle on the sides $A B , B C$ and $C A$ respectively, then $l _ { 1 } ^ { 2 } + l _ { 2 } ^ { 2 } + l _ { 3 } ^ { 2 }$ equals :
(1) $\frac { 1 } { 5 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 3 }$

Let $\overrightarrow { \mathrm { a } } = \hat { i } + 2 \hat { j } + 3 \hat { k } , \overrightarrow { \mathrm {~b} } = 2 \hat { i } + 3 \hat { j } - 5 \hat { k }$ and $\overrightarrow { \mathrm { c } } = 3 \hat { i } - \hat { j } + \lambda \hat { k }$ be three vectors. Let $\overrightarrow { \mathrm { r } }$ be a unit vector along $\vec { b } + \vec { c }$. If $\vec { r } \cdot \vec { a } = 3$, then $3 \lambda$ is equal to: (1) 21 (2) 30 (3) 25 (4) 27

Let $P$ and $Q$ be the points on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ which are at a distance of 6 units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha, \beta, \gamma)$, then $\alpha^2 + \beta^2 + \gamma^2$ is:
(1) 26
(2) 36
(3) 18
(4) 24

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
(1) $\frac { \pi } { 6 }$
(2) $\cos ^ { - 1 } \left( \frac { 7 } { 18 } \right)$
(3) $\cos ^ { - 1 } \left( \frac { 1 } { 18 } \right)$
(4) $\frac { \pi } { 3 }$

Let $L_1: \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i} - \hat{j} + 2\hat{k})$, $\lambda \in R$, $L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(3\hat{i} + \hat{j} + p\hat{k})$, $\mu \in R$ and $L_3: \vec{r} = \delta(l\hat{i} + m\hat{j} + n\hat{k})$, $\delta \in R$ 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
(1) $(-1, 7, 4)$
(2) $(-1, -7, 4)$
(3) $(1, 7, -4)$
(4) $(1, -7, 4)$

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 R be the point $( 2,5 , - 1 )$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda ^ { 2 } = 14 K$, then $K$ is equal to:
(1) 36
(2) 81
(3) 72
(4) 18

If the shortest distance between the lines $\frac { x - \lambda } { 3 } = \frac { y - 2 } { - 1 } = \frac { z - 1 } { 1 }$ and $\frac { x + 2 } { - 3 } = \frac { y + 5 } { 2 } = \frac { z - 4 } { 4 }$ is $\frac { 44 } { \sqrt { 30 } }$, then the largest possible value of $| \lambda |$ is equal to $\_\_\_\_$

A line with direction ratio $2,1,2$ meets the lines $\mathrm { x } = \mathrm { y } + 2 = \mathrm { z }$ and $\mathrm { x } + 2 = 2 \mathrm { y } = 2 \mathrm { z }$ respectively at the point P and Q . if the length of the perpendicular from the point $( 1,2,12 )$ to the line PQ is $l$, then $l ^ { 2 }$ is

Let a line passing through the point $(-1, 2, 3)$ intersect the lines $L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$ at $N(a, b, c)$. Then the value of $\frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2}$ equals $\underline{\hspace{1cm}}$.

The square of the distance of the image of the point $( 6,1,5 )$ in the line $\frac { x - 1 } { 3 } = \frac { y } { 2 } = \frac { z - 2 } { 4 }$, from the origin is $\_\_\_\_$

Let in a $\triangle ABC$, the length of the side $AC$ 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 $\triangle ABC$ is:
(1) 17
(2) 21
(3) 56
(4) 42