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 $\vec { a } = 2 \hat { i } + \alpha \hat { j } + \hat { k } , \vec { b } = - \hat { i } + \hat { k } , \vec { c } = \beta \hat { j } - \hat { k }$, where $\alpha$ and $\beta$ are integers and $\alpha \beta = - 6$. Let the values of the ordered pair ( $\alpha , \beta$ ), for which the area of the parallelogram of diagonals $\vec { a } + \vec { b }$ and $\vec { b } + \vec { c }$ is $\frac { \sqrt { 21 } } { 2 }$, be ( $\alpha _ { 1 } , \beta _ { 1 }$ ) and $\left( \alpha _ { 2 } , \beta _ { 2 } \right)$. Then $\alpha _ { 1 } ^ { 2 } + \beta _ { 1 } ^ { 2 } - \alpha _ { 2 } \beta _ { 2 }$ is equal to
(1) 19
(2) 17
(3) 24
(4) 21

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 $P Q R$ be a triangle with $R ( - 1,4,2 )$. Suppose $M ( 2,1,2 )$ is the mid point of $P Q$. The distance of the centroid of $\triangle P Q R$ from the point of intersection of the line $\frac { x - 2 } { 0 } = \frac { y } { 2 } = \frac { z + 3 } { - 1 }$ and $\frac { x - 1 } { 1 } = \frac { y + 3 } { - 3 } = \frac { z + 1 } { 1 }$ is
(1) 69
(2) 9
(3) $\sqrt { 69 }$
(4) $\sqrt { 99 }$

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) $(1, -7, 4)$

For $\lambda > 0$, let $\theta$ be the angle between the vectors $\vec { a } = \hat { i } + \lambda \hat { j } - 3 \hat { k }$ and $\vec { b } = 3 \hat { i } - \hat { j } + 2 \hat { k }$. If the vectors $\vec { a } + \vec { b }$ and $\vec { a } - \vec { b }$ are mutually perpendicular, then the value of $( 14 \cos \theta ) ^ { 2 }$ is equal to
(1) 50
(2) 40
(3) 25
(4) 20

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 :
(1) 16
(2) 20
(3) 14
(4) 18

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
(1) $\gamma \sqrt { 1 - \sin ^ { 2 } \phi \cos ^ { 2 } \theta }$
(2) $\gamma \sqrt { 1 - \sin ^ { 2 } \theta \cos ^ { 2 } \phi }$
(3) $\gamma \sqrt { 1 + \cos ^ { 2 } \phi \sin ^ { 2 } \theta }$
(4) $\gamma \sqrt { 1 + \cos ^ { 2 } \theta \sin ^ { 2 } \phi }$

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

Consider the line $L$ 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 L along the line $\frac { 3 x - 11 } { 2 } = \frac { 3 y - 11 } { 1 } = \frac { 3 z - 19 } { 2 }$ is equal to
(1) 6
(2) 5
(3) 4
(4) 3

Let 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 P from the line $4 x = 2 y = z$ is
(1) $\frac { 5 \sqrt { 14 } } { 7 }$
(2) $\frac { 3 \sqrt { 14 } } { 7 }$
(3) $\frac { \sqrt { 14 } } { 7 }$
(4) $\frac { 6 \sqrt { 14 } } { 7 }$

Let $\vec { a } = 9 \hat { i } - 13 \hat { j } + 25 \hat { k } , \vec { b } = 3 \hat { i } + 7 \hat { j } - 13 \hat { k }$ and $\vec { c } = 17 \hat { i } - 2 \hat { j } + \hat { k }$ be three given vectors. If $\vec { r }$ is a vector such that $\vec { r } \times \vec { a } = ( \vec { b } + \vec { c } ) \times \vec { a }$ and $\vec { r } \cdot ( \vec { b } - \vec { c } ) = 0$, then $\frac { | 593 \vec { r } + 67 \vec { a } | ^ { 2 } } { ( 593 ) ^ { 2 } }$ 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

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 :
(1) $\sqrt { 17 }$
(2) $\sqrt { 15 }$
(3) $\sqrt { 14 }$
(4) $\sqrt { 13 }$

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

Let $\hat { a }$ be a unit vector perpendicular to the vector $\overrightarrow { \mathrm { b } } = \hat { i } - 2 \hat { j } + 3 \hat { k }$ and $\overrightarrow { \mathrm { c } } = 2 \hat { i } + 3 \hat { j } - \hat { k }$, and makes an angle of $\cos ^ { - 1 } \left( - \frac { 1 } { 3 } \right)$ with the vector $\hat { i } + \hat { j } + \hat { k }$. If $\hat { a }$ makes an angle of $\frac { \pi } { 3 }$ with the vector $\hat { i } + \alpha \hat { j } + \hat { k }$, then the value of $\alpha$ is :
(1) $\sqrt { 6 }$
(2) $- \sqrt { 6 }$
(3) $- \sqrt { 3 }$
(4) $\sqrt { 3 }$

Let the position vectors of the vertices $A , B$ and $C$ of a tetrahedron $A B C D$ be $\hat { \mathbf { i } } + 2 \hat { \mathbf { j } } + \hat { \mathbf { k } } , \hat { \mathbf { i } } + 3 \hat { \mathbf { j } } - 2 \hat { k }$ and $2 \hat { i } + \hat { j } - \hat { k }$ respectively. The altitude from the vertex $D$ to the opposite face $A B C$ meets the median line segment through $A$ of the triangle $A B C$ at the point $E$. If the length of $A D$ is $\frac { \sqrt { } \overline { 110 } } { 3 }$ and the volume of the tetrahedron is $\frac { \sqrt { 805 } } { 6 \sqrt { 2 } }$, then the position vector of $E$ is
(1) $\frac { 1 } { 12 } ( 7 \hat { \mathbf { i } } + 4 \hat { \mathbf { j } } + 3 \hat { k } )$
(2) $\frac { 1 } { 2 } ( \hat { i } + 4 \hat { j } + 7 \hat { k } )$
(3) $\frac { 1 } { 6 } ( 12 \hat { i } + 12 \hat { j } + \hat { k } )$
(4) $\frac { 1 } { 6 } ( 7 \hat { \mathrm { i } } + 12 \hat { \mathrm { j } } + \hat { \mathrm { k } } )$

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 :
(1) 148
(2) 136
(3) 144
(4) 140

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 :
(1) 25
(2) 19
(3) 29
(4) 27

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 $\frac { \mathrm { m } } { \mathrm { n } }$, where $\mathrm { m } , \mathrm { n }$ are coprime numbers, then $\mathrm { m } + \mathrm { n }$ is equal to :
(1) 21
(2) 9
(3) 14
(4) 6

Let $\overrightarrow{\mathrm{a}} = \hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{\mathrm{b}} = 2\hat{i} + 7\hat{j} + 3\hat{k}$. Let $\mathrm{L}_1: \overrightarrow{\mathrm{r}} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda\overrightarrow{\mathrm{a}},\ \lambda \in \mathbf{R}$ and $\mathrm{L}_2: \overrightarrow{\mathrm{r}} = (\hat{j} + \hat{k}) + \mu\overrightarrow{\mathrm{b}},\ \mu \in \mathbf{R}$ be two lines. If the line $\mathrm{L}_3$ passes through the point of intersection of $\mathrm{L}_1$ and $\mathrm{L}_2$, and is parallel to $\vec{a} + \vec{b}$, then $\mathrm{L}_3$ passes through the point:
(1) $(5, 17, 4)$
(2) $(2, 8, 5)$
(3) $(8, 26, 12)$
(4) $(-1, -1, 1)$

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
(1) 9
(2) 12
(3) 7
(4) 8

Let the position vectors of three vertices of a triangle be $4\vec{p} + \vec{q} - 3\vec{r}$, $-5\vec{p} + \vec{q} + 2\vec{r}$ and $2\vec{p} - \vec{q} + 2\vec{r}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p} + \vec{q} + \vec{r}}{4}$ and $\alpha\vec{p} + \beta\vec{q} + \gamma\vec{r}$ respectively, then $\alpha + 2\beta + 5\gamma$ is equal to:
(1) 3
(2) 4
(3) 1
(4) 6

Let $\mathrm { A } ( x , y , z )$ be a point in $xy$-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 $\frac { 9 \sqrt { 2 } } { 2 }$,
(1) both are true
(2) only (S2) is true
(3) only (S1) is true
(4) both are false