The direction ratios of the line bisecting the angle between the $X$-axis and the line having direction ratios $(3,-1,5)$ are

If the plane $-4 x-2 y+2 z+\alpha=0$ is at a distance of two units from the plane $2 x+y-z+1=0$, then the product of all the possible values of $\alpha$ is

The equation of the locus of a point whose distance from $X Y$-plane is twice its distance from $Z$-axis is

If $\alpha$ is the angle between any two diagonals of a cube and $\beta$ is the angle between a diagonal of a cube and a diagonal of its face, which intersects this diagonal of the cube, then $\cos \alpha+\cos ^2 \beta=$

If the angle between the planes $a x-y+3 z=2 a$ and $3 x+a y+z=3 a$ is $\frac{\pi}{3}$, then the direction ratio of the line perpendicular to the plane $(a+2) x+(a-4) y+2 a z=a$ are

The number of values of ' $k$ ' for which the points $(-4,9, k),(-1,6, k),(0,7,10)$ from right-angled isosceles triangle is

A line makes angles $60^{\circ}, 45^{\circ}, \theta$ with positive $X, Y, Z$ axes respectively. If $\theta$ is an acute angle, then $\tan \theta=$

If the foot of the perpendicular drawn from the point $(2,0,-3)$ to the plane $\pi$ is $(1,-2,0)$ and the equation of the plane $\pi$ is $a x+b y-3 z+d=0$, then $a+b+d=$

Let $\pi_1$ be the plane determined by the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$. $\hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\pi_2$ be the plane determined by the vectors $\hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{k}}-\hat{\mathbf{i}}$. Let $\mathbf{a}$ be a non-zero vector parallel to the line of intersection of the planes $\pi_1$ and $\pi_2$. If $\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$, then the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$ is

If $m: n$ is the ratio in which the point $\left(\frac{8}{5},-\frac{1}{5}, \frac{8}{5}\right)$ divides the segment joining the points $(2, p, 2)$ and $(p,-2, p)$, where $p$ is an integer than $\frac{3 m+n}{3 n}=$

If $(\alpha, \beta \gamma)$ is the foot of the perpendicular drawn from a point $(-1,2,-1)$ to the line joining the points $(2,-1,1)$ and ( $1,1-2$ ), then $\alpha+\beta+\gamma=$

If $A(2,1,-1), B(6,-3,2), C(-3,12,4)$ are the vertices of a $\triangle A B C$ and the equation of the plane containing the $\triangle A B C$ is $53 x+b y+c z+d=0$, then $\frac{d}{b+c}=$

Let $A$ be a point having position vector $\hat{\mathbf{i}}-3 \hat{\mathbf{j}}$ and $\mathbf{r}=(\hat{\mathbf{i}}-3 \hat{\mathbf{j}})+t(\hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ be a line. If $P$ is a point on this line and is at a minimum distance from the plane $\mathbf{r} .(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})=0$, then the equation of the plane through $P$ and perpendicular to $A P$, is

If $L$ is a line common to the planes $3 x+4 y+7 z=1$, $x-y+z=5$, then the direction ratios of the line $L$ are

If the points $(1,1, \lambda)$ and $(-3,0,1)$ are equidistant from the plane $3 x+4 y-12 z+13=0$, then the values of $\lambda$ are

The shortest distance between the lines

$ \begin{aligned} & \mathbf{r}=(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+t(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and } \\ & \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})+s(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \text { is } \end{aligned} $

If $A(0,3,4), B(1,5,6), C(-2,0,-2)$ are the vertices of a $\triangle A B C$ and the bisector of angle $A$ meets the side $B C$ at $D$, then $A D=$

If the direction cosines of two lines satisfy the equation $2 l+m-n=0, l^2-2 m^2+n^2=0$ and $\theta$ is the angle between the lines, then $\cos \theta=$

If the equation of the plane passing through the points $(2,1,2),(1,2,1)$ and perpendicular to the plane $2 x-y+2 z=1$ is $a x+b y+c z+d=0$, then $\frac{a+b}{c+d}=$

If the circumcenter of the triangle formed by the points $(1,2,3),(3,-1,5)$ and $(4,0,-3)$ is $(\alpha, \beta, \gamma)$, then $|\alpha|+|\beta|=$

If $\theta$ is the acute angle between the two lines whose direction cosines are connected by the relations $l+m+n=0$ and $2 l m+2 n l-m n=0$, then $\cos \theta=$

If the foot of the perpendicular drawn from the point $(1,0,-2)$ to the plane $\pi$ is $(2,0,-1)$ and the equation of the plane $\pi$ is $a x+b y+c z=2$, then $a^2+b^2+c^2=$

If $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$ are points in a plane, then the vector equation of the line passing through the centroids of $\triangle A B D$ and $\triangle A C D$ is

In a $\triangle A B C$, if the mid-points of sides $A B, B C$ and $C A$ are $(3,0,0),(0,4,0)$ and $(0,0,5)$ respectively, then $A B^2+B C^2+C A^2=$

If $l, m, n$ and $a, b, c$ are direction cosines of two lines, then

If $(2,-1,3)$ is the foot of the perpendicular drawn from the origin to a plane, then the equation of that plane is

$A(1,2,3), B(2,3,1)$ and $C(3,1,2)$ are three points. If the point $P$ divides $A B$ in the ratio $1: 2$ and the point $Q$ divides $B C$ in the ratio $-2: 3$, then the distance between $P$ and $Q$ is

If the image of the point $(1,-2,1)$ with respect to the line passing through the points $B(1,1,2)$ and $C(2,2,1)$ is $(l, m, n)$, then $l^2+m^2+n^2=$

A plane $\pi$ passing through the point $(1,1,1)$ is perpendicular to the line joining the points $(6,3,2)$ and $(1,-4,-9)$. If $a x+b y+c z-23=0$ is the equation of the plane $\pi$, then $a+b-c=$

The point of intersection of the line passing through the point $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and the plane passing through the points $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}, 2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+2 \hat{\mathbf{k}}$ is

A plane $\pi$ passing through the point $3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ is parallel to the plane which passes through the point $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and perpendicular to the vector $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$. Then, the cartesian equation of $\pi$ is

Let the direction cosines of two lines satisfy the equations $3 l+2 m+n=0$ and $2 m n-3 n l+5 l m=0$. If $\theta$ is the angle between these two lines, then $\cos \theta=$

$(1,-2,1)$ is a point on a plane $\pi$ and $\pi$ is parallel to the plane $x-y-z=0$. If the equation of $\pi$ is $a x+b y+c z-2=0$, then $b-2 c=$

If $M$ is the foot of the perpendicular drawn from $P($ -1,2,-1 ) to the plane passing through the point $A(3,-2,1)$ and perpendicular to the vector $4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$, then the length of $P M$ is

If $A=(1,-1,2), B=(3,4,-2), C=(0,3,2)$ and $D=(3$, $5,6)$, then the angle between the lines $\mathbf{A B}$ and $\mathbf{C D}$ is

Consider the following statements:

Assertion (A) : The direction ratios of a line $L_1$ are 2,5, 7 and the direction ratios of another line $L_2$ are $\frac{4}{\sqrt{19}}$, $\frac{10}{\sqrt{19}}, \frac{14}{\sqrt{19}}$. Then, the lines $L_1, L_2$ are parallel.

Reason : ( $\mathbf{R}$ ) If the direction ratios of a line $L_1$ are $a_1, b_1, c_1$ the direction ratios of a line $L_2$ are $a_2, b_2, c_2$ and $a_1 a_2+b_1 b_2+c_1 c_2=0$, then the lines of $L_1, L_2$ are parallel.

A line $L$ is parallel to both the planes $2 x+3 y+z=1$ and $x+3 y+2 z=2$. If line $L$ makes an angle $\alpha$ with the positive direction of $X$-axis, then $\cos \alpha=$

If $\mathbf{r}=(2-\lambda+\mu) \hat{\mathbf{i}}+(1-\mu) \hat{\mathbf{j}}+(2-3 \lambda+2 \mu) \hat{\mathbf{k}}$ is the vector equation of a plane, then the equivalent cartesian equation of the plane is

Let $\pi_1$ be a plane passing through the point $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and perpendicular to the vector $-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$. Let the line $L$ passing through the points $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ be a normal to the plane $\pi_2$. If the angle between the planes $\pi_1$ and $\pi_2$ is $\theta$, then $\cos \theta=$

Let $A=(1,2,0), B=(2,0,-1), C=(0,-2,3)$ and $D=(-1,2,-3)$ be four points in the space. Let $G_1$ be the centroid of $\triangle A B C$ and $G_2$ be the centroid of tetrahedron $A B C D$. If $P$ divides, $G_1 G_2$ in the ratio $4: 3$ internally, then $P=$

If the d.r.'s of two lines are connected by the relations $a-b+c=0, a^2-b^2+2 c^2=0$ and $\theta$ is the angle between these lines, then $\cos \theta=$

If $l, m$ and $n$ are the d.c.'s of a normal to the plane passing through the points $(0,1,2)$, $(3,0,2)$ and $(4,5,0)$, then $|I|+|m|+|n|=$

Let $L$ be a line passing through the points $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+6 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$. Let $P$ be a plane passing through $-5 \hat{\mathbf{i}}+19 \hat{\mathbf{j}}-14 \hat{\mathbf{k}}$ and parallel to the vectors $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$. If $L$ meets the plane $P$ at a point $A$, then the position vector of $A$, is

If $\mathbf{r} \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=5, \mathbf{r} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=7$ are two planes and $(16,-9,0)$ is a point common to both the planes, then the vector equation of the line of intersection of the planes is $\mathbf{r}=$

$A(1,1,1), B(1,-4,3), C(2,-2,0)$ and $D(8,1,4)$ are the vertices of a tetrahedron. $G_1, G_2, G_3$ and $G_4$ are the centroids of the faces $A B C, B C D, C D A$ and $D A B$. Then, the centroid of the tetrahedron having $G_1, G_2, G_3$ and $G_4$ as its vertices is

Let $A(2,3,-1), B(4,1,0), C(-1,-1,1)$ be the vertices of a $\triangle A B C$. Let $D$ be the point where the bisector of $B A C$ meet the side $B C$. Then, the direction ratios of $A D$ are

If a plane passing through the points $(2,3,0),(0,-5,2)$ and ( $-2,0,3$ ) meets the $X, Y$ and $Z$-axes in $A, B$ and $C$ respectively, then $A=$

The point which lies on the plane passing through the point $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ is

If the angle between the planes $\mathbf{r} \cdot(11 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}})=7$ and $\mathbf{r} \cdot(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})=5$ is $\frac{\pi}{2}$, then $\alpha=$

$A(27,-243,81)$ is a point in space, $B, C$ and $D$ are images of $A$ with respect to $X Y, Y Z$ and $Z X$ planes respectively. If the centroid of the $\triangle B C D$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma=$

Let $A(2,5,7)$ be the image of the point $B(1,-2,3)$ with respect to a plane $\pi$. Let $C$ be the point where $A B$ meets the plane $\pi$. Let $D=(2,1,6)$. Then, the direction cosines of $C D$ are

If a plane $x+y+z-5=0$ intersects the line joining $A(1,1,1)$ and $B(2,2,2)$ at $P$, then $A P: P B=$

Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$. Let $-7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$ be the position vector of a point $P$ on $L$ such that $|\mathbf{A P}|=12$. Then, the position vector of $\mathbf{A}$ can be

A bisector of the angle between the normals of the planes $4 x+3 y=5$ and $x+2 y+2 z=4$ is along the vector

If $A(1,2,3), B(2,-3,1), C(3,2,-1)$ are three vertices of a tetrahedron $A B C D$ and $G\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ is its centroid, then the point which divides $G D$ in the ratio $1: 2$ is

Let $D$ be the foot of the perpendicular drawn from the point $A(2,0,3)$ to the line joining the points $B(0,4,1)$ and $C(-2,0,4)$. Then, the ratio in which $D$ divides $B C$ is

Let $6 x-3 y+2 z-6=0$ be the given plane. If $a, b$ and $c$ are the intercepts made by the plane on $X, Y$ and $Z$-axes, respectively; $l, m$ and $n$ are the direction cosines of a normal drawn to the plane and $p$ is the perpendicular distance from the origin to the plane, then $|a l+b m+c n|=$

Let a plane $P$ has the points $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$. Let $L$ be the line through the point $A$ and parallel to the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If the plane $P$ and line $L$ intersect at a point $B(0,3,2)$ and the distance from $A$ to $B$ is 3 units, then equations of the normal to the plane $P$ through $A$ are

Let $\pi_1^{\prime}$ be the plane passing through the point $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and perpendicular to the vector $a \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $\pi_2$ be the plane passing through the point $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and perpendicular to the vector $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\theta$ is the angle between the planes $\pi_1$ and $\pi_2$ and $\cos \theta=-\sqrt{\frac{3}{7}}$, then the integral value of $a$ is

If the points $A(1,3,5), B(2,4,6), C(4,5, k)$ form a right angled triangle then the number of possible values of $k$ is

Let $A=(3,4,0), B=(4,4,4), C=(-6,2,3)$ and $D=(1,1,2)$. If $\theta$ is the acute angle between the lines $A B$ and $C D$, then $\cos \theta=$

A plane containing two lines whose direction ratios are $(-1,2,1)$ and $(1,3,2)$ passes through the point $(2,1, k)$. If this plane also passes through the point $(3,-1,4)$, then $k=$

If $P$ is a point on the line parallel to the vector $2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$ and passing through the point $A$ whose position vector is $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $A P=21$, then the position vector of $P$ can be

The cartesian equation of the plane passing through the point $(1,-2,3)$ and perpendicular to the vector $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$, is

Let $A(1,2,3), B(-1,4,6), C(0,-6,4)$ and $D(1,1,1)$ be the vertices of a tetrahedron, $G$ be its centroid and $G_1$ be the centroid of its face $B C D$. Then, $\frac{A G_1}{A G}=$

If a line $L$ is common to the planes $x-y+z+2=0$ and $2 x+y-2 z+5=0$ then the direction cosines of the line $L$ are

Let the foot of the perpendicular drawn from the point $(1,2,3)$ to a plane be $(-1,3,-2)$. Then, the perpendicular distance from the origin to the plane is

If $A(4,7,8), B(2,3,4)$ and $C(2,5,7)$ are the vertices of $\triangle A B C$, then the length of the internal bisector of the angle $A$ is

For scalars $\lambda, \mu$ if the vector equation of a plane is $\mathbf{r}=(2+3 \lambda-\mu) \hat{\mathbf{i}}+(1-2 \lambda+3 \mu) \hat{\mathbf{j}}+(-2+2 \lambda+\mu) \hat{\mathbf{k}}$, then its Cartesian equation is

The position vectors of the points $A$ and $B$ are respectively $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If the points $P$ and $Q$ are respectively the orthogonal projections of $A$ and $B$ on the plane $x+y+z=3$, then $P Q=$

If $A(4,3,2), B(5,4,6), C(-1,-1,5)$ are the vertices of a triangle, then the coordinates of the point in which the bisector of the angle $A$ meet the side $B C$ is

Assertion (A) The direction ratios of line $L_1$ are 2, 5, 7 and those of line $L_2$ are $\frac{4}{\sqrt{19}}, \frac{10}{\sqrt{19}}, \frac{14}{\sqrt{19}}$. The lines $L_1, L_2$ are parallel.

$\boldsymbol{\operatorname { R e a s o n }}(R)$ The direction ratios of a line $L_1$ are $a_1, b_1, c_1$ and those of another line $L_2$ are $a_2, b_2, c_2$. The lines $L_1$ and $L_2$ are parallel if $a_1 a_2+b_1 b_2+c_1 c_2=0$

The correct option among the following is

If $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$ lies in the plane $a x+b y+z=7$, then $a+b=$

The shortest distance between the skew-lines $\mathbf{r}=(-\hat{\mathbf{i}}+3 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})$ and $\mathbf{r}=(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+s(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ is

$\Pi_1, \Pi_2, \Pi_3$ are three planes which are respectively parallel to the $Y Z, Z X$ and $X Y$ planes at distances $a, b$ and $c$ forming a rectangular parallelopiped. $d_1$ is a diagonal of the face of $X Y$-plane not passing through the origin and $d_2$ is a diagonal of the plane $\Pi_2$ coterminous with $d_1$. If none of the coordinates of the vertices of the parallelopiped are negative, then the angle between $d_1$ and $d_2$ is

The obtuse angle between the lines whose direction ratios are determined by the equations $a+b+c=0$, $2 a b+2 a c-b c=0$ is

A plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle A B C$ is $(2,3,5)$. Then, the equation of that plane is

Let $\Pi$ be a plane containing the points $(0,-5,-1),(1,-2,5),(-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$. Then the length of the projection of the unit normal vector to the plane $\Pi$ on the line $L$ is

If the line passing through the points $(a, 2,-4)$ and $(5,3, b)$ crosses the $Z X$-plane at the point $(-a+2 b, 0, a+b)$, then $14 a+7 b$

The direction cosines of the normal to the plane containing the lines having direction ratios $1,2,1$ and 4,5, -3 are

The foot of the perpendicular drawn from the point $(1,1,1)$ to the plane $\pi_1$ is $(1,3,5)$. If $(2,2,-1),(3,4,2)$, $(3,3,0)$ are three points on the plane $\pi_2$, then the angle between the planes $\pi_1$ and $\pi_2$ is

The position vector of a point $P$ is $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{a}=-\hat{\mathbf{i}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are two vectors which determine a plane $\pi$. The equation of a line through $P$ normal to $\mathbf{b}$ and lying on the plane $\pi$ is

The shortest distance between the line $\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})$ and the plane $\mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5$ is

If the points $A(-1,0,7), B(3,2, t), C(5, k,-2)$ are collinear, then the ratio in which the point $P(t, k-2 t, t+k)$ divides the line segment $B C$ is

The direction cosines $l, m, n$ of two lines are satisfying $3 l+m+5 n=0$ and $6 m n-2 n l+5 l m=0$. If $\theta$ is the angle between those lines then $|\cos \theta|=$

A tetrahedron has vertices $O(0,0,0), A(1,2,1)$, $B(2,1,3), C(-1,1,2)$. If $\theta$ is the angle between the faces $O A B$ and $A B C$, then $\cos \theta=$