3D Geometry
Consider a line $\mathrm{L}$ passing through the points $\mathrm{P}(1,2,1)$ and $\mathrm{Q}(2,1,-1)$. If the mirror image of the point $\mathrm{A}(2,2,2)$ in the line $\mathrm{L}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+6 \gamma$ is equal to __________.
Explanation:
$\begin{aligned} & A(2,2,2) \\ & P Q: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{-2} \end{aligned}$
General point,
$(k+1,-k+2,-2 k+1)$
$\begin{aligned} & \overrightarrow{O A}=(k+1-2) \hat{i}+(-k+2-2) \hat{j}+(-2 k+1-2) \hat{k} \\ & \overrightarrow{O A}=(k-1) \hat{i}-k \hat{j}+(-2 k-1) \hat{k} \\ & \overrightarrow{P Q}=1 \hat{i}-\hat{j}-2 \hat{k} \\ & \overrightarrow{O A} \cdot \overrightarrow{P Q}=0 \\ & (k-1)+k+2(2 k+1)=0 \\ & k-1+k+4 k+2=0 \\ & 6 k+1=0 \\ & k=\frac{-1}{6} \\ & 0\left(\frac{-1}{6}+1, \frac{+1}{6}+2,-2\left(\frac{-1}{6}\right)+1\right) \\ & 0\left(\frac{5}{6}, \frac{13}{6}, \frac{-8}{6}\right) \\ & 0\left(\frac{5}{6}, \frac{13}{6}, \frac{8}{6}\right)=\left(\frac{\alpha+2}{2}, \frac{\beta+2}{2}, \frac{\gamma+2}{2}\right) \end{aligned}$
$\begin{array}{l|l} \alpha+2=\frac{10}{6} & \beta+2=\frac{26}{6} \\ \alpha=\frac{10}{6}-2 & \beta=\frac{26-12}{6} \\ \alpha=\frac{-2}{6} & \beta=\frac{14}{6} \\ \alpha =-\frac{1}{3} & \beta=\frac{7}{3} \end{array}$
$\begin{aligned} & \gamma+2=\frac{16}{6} \\ & \gamma=\frac{16-12}{6} \\ & \gamma=\frac{4}{6} \\ & \Rightarrow \alpha+\beta+6 \gamma \\ & \Rightarrow \frac{-1}{3}+\frac{7}{3}+6 \times \frac{4}{6} \\ & \Rightarrow 2+4=6 \end{aligned}$
$ \begin{aligned} & \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\ & \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k}) \end{aligned} $
intersect $\mathrm{L}_1$ and $\mathrm{L}_2$ at $\mathrm{P}$ and $\mathrm{Q}$ respectively. If $(\alpha, \beta, \gamma)$ is the mid point of the line segment $\mathrm{PQ}$, then $2(\alpha+\beta+\gamma)$ is equal to ____________.
Explanation:
$P \equiv(\lambda+1,-\lambda+2, \lambda+3)$
$Q \equiv(\mu+4, \mu+5,-\mu+6)$
$\overrightarrow{P Q}=(\mu-\lambda+3, \mu+\lambda+3,-\mu-\lambda+3)$
$\overrightarrow{P Q} \cdot \vec{b}_1=0 \Rightarrow 3 \lambda+\mu=3$ ..........(i)
$\overrightarrow{P Q} \cdot \vec{b}_2=0 \Rightarrow 3 \mu+\lambda=3$ ..........(ii)
From (i) and (ii),
$ \begin{aligned} & P \equiv\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) \& Q \equiv\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right) \\\\ & \alpha=\frac{5}{2}, \beta=\frac{4}{2}, \gamma=\frac{12}{2} \\\\ & 2(\alpha+\beta+\gamma)=21 \end{aligned} $
A line passes through $A(4,-6,-2)$ and $B(16,-2,4)$. The point $P(a, b, c)$, where $a, b, c$ are non-negative integers, on the line $A B$ lies at a distance of 21 units, from the point $A$. The distance between the points $P(a, b, c)$ and $Q(4,-12,3)$ is equal to __________.
Explanation:
$\begin{aligned} & \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \\ & \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \\ & \left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) \\ & =(22,0,7)=(a, b, c) \\ & \therefore \sqrt{324+144+16}=22 \end{aligned}$
Let $\mathrm{Q}$ and $\mathrm{R}$ be the feet of perpendiculars from the point $\mathrm{P}(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle \mathrm{QPR}$ is a right angle, then $12 a^2$ is equal to _________.
Explanation:
$\begin{aligned} & \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\ & \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \\ & \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\ & a=r+a-r=0 \\ & 2 a=2 r \rightarrow a=r \\ & \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\ & a-k-a-k=0 \Rightarrow k=0 \\ & A s, P Q \perp P R \\ & (a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 \\ & a=1 \text { or }-1 \\ & 12 a^2=12 \end{aligned}$
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 __________.
Explanation:
$\begin{aligned} & \mathrm{M}(3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2 \\ & \mathrm{~N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1 \end{aligned}$
$\begin{aligned} & \frac{3 \lambda+2}{-3 \mu-1}=\frac{2 \lambda}{-2 \mu}=\frac{-2 \lambda-4}{4 \mu-2} \\ & 3 \lambda \mu+2 \mu=3 \lambda \mu+\lambda \\ & 2 \mu=\lambda \\ & 2 \lambda \mu-\lambda=\lambda \mu+2 \mu \\ & \lambda \mu=\lambda+2 \mu \\ &\Rightarrow \lambda \mu=2 \lambda \end{aligned}$
$\begin{aligned} \Rightarrow \quad & \mu=2 \quad(\lambda \neq 0) \\ \therefore \quad & \lambda=4 \\ & \alpha+\beta+\gamma=14 \\ & a+b+c=-1 \\ & \frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}=196 \end{aligned}$
If $\mathrm{d}_1$ is the shortest distance between the lines $x+1=2 y=-12 z, x=y+2=6 z-6$ and $\mathrm{d}_2$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$, then the value of $\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}$ is :
Explanation:
$\mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$
$\mathrm{d}_1=$ shortest distance between $\mathrm{L}_1 ~\& \mathrm{~L}_2$
$\begin{aligned} & =\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}\right| \\ & d_1=2 \\ & \mathrm{~L}_3: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}_4: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3} \\ & \mathrm{~d}_2=\text { shortest distance between } \mathrm{L}_3 ~\& \mathrm{~L}_4 \\ & \mathrm{~d}_2=\frac{12}{\sqrt{3}} \text { Hence } \\ & =\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16 \end{aligned}$
Let O be the origin, and M and $\mathrm{N}$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $\mathrm{MN}$ is the shortest distance between the given lines. Then $\overrightarrow{O M} \cdot \overrightarrow{O N}$ is equal to _________.
Explanation:
$\mathrm{L}_1: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad \operatorname{drs}(4,1,3)=\mathrm{b}_1$
$\begin{aligned} & \mathrm{M}(4 \lambda+5, \lambda+4,3 \lambda+5) \\ & \mathrm{L}_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu \\ & \mathrm{N}(12 \mu-8,5 \mu-2,9 \mu-11) \\ & \overrightarrow{\mathrm{MN}}=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16) \quad \text{..... (1)} \end{aligned}$
Now
$\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{array}\right|=-6 \hat{\mathrm{i}}+8 \hat{\mathrm{k}} \quad \text{.... (2)}$
Equation (1) and (2)
$\therefore \frac{4 \lambda-12 \mu+13}{-6}=\frac{\lambda-5 \mu+6}{0}=\frac{3 \lambda-9 \mu+16}{8}$
I and II
$\lambda-5 \mu+6=0 \quad \text{.... (3)}$
I and III
$\lambda-3 \mu+4=0 \quad \text{.... (4)}$
Solve (3) and (4) we get
$\begin{aligned} \lambda= & -1, \mu=1 \\ \therefore \quad & \mathrm{M}(1,3,2) \\ & \mathrm{N}(4,3,-2) \\ \therefore \quad & \overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{ON}}=4+9-4=9 \end{aligned}$
A line with direction ratios $2,1,2$ meets the lines $x=y+2=z$ and $x+2=2 y=2 z$ respectively at the points $\mathrm{P}$ and $\mathrm{Q}$. If the length of the perpendicular from the point $(1,2,12)$ to the line $\mathrm{PQ}$ is $l$, then $l^2$ is __________.
Explanation:
Let $\mathrm{P}(\mathrm{t}, \mathrm{t}-2, \mathrm{t})$ and $\mathrm{Q}(2 \mathrm{~s}-2, \mathrm{~s}, \mathrm{~s})$
D.R's of PQ are 2, 1, 2
$\begin{aligned} & \frac{2 \mathrm{~s}-2-\mathrm{t}}{2}=\frac{\mathrm{s}-\mathrm{t}+2}{1}=\frac{\mathrm{s}-\mathrm{t}}{2} \\ & \Rightarrow \mathrm{t}=6 \text { and } \mathrm{s}=2 \\ & \Rightarrow \mathrm{P}(6,4,6) \text { and } \mathrm{Q}(2,2,2) \\ & \mathrm{PQ}: \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-2}{2}=\lambda \end{aligned}$
Let $\mathrm{F}(2 \lambda+2, \lambda+2,2 \lambda+2)$
$\mathrm{A}(1,2,12)$
$\overrightarrow{\mathrm{AF}} \cdot \overrightarrow{\mathrm{PQ}}=0$
$\therefore \lambda=2$
So $F(6,4,6)$ and $A F=\sqrt{65}$
The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $\mathrm{P}$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$, then $14 l^2$ is equal to __________.
Explanation:
$\begin{aligned} & \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-7}{8}=\lambda \\ & \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} \\ & \Rightarrow \lambda+2=4 \mathrm{k}-3 \\ & -\lambda=3 \mathrm{k}-2 \\ & \Rightarrow \mathrm{k}=1, \lambda=-1 \\ & 8 \lambda+7=\mathrm{k}-2 \\ & \therefore \mathrm{P}=(1,1,-1) \end{aligned}$
Projection of $2 \hat{i}-2 \hat{k}$ on $2 \hat{i}+3 \hat{j}+\hat{k}$ is
$\begin{aligned} & =\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}} \\ & \therefore l^2=8-\frac{4}{14}=\frac{108}{14} \\ & \Rightarrow 14 l^2=108 \end{aligned}$
Let $\gamma \in \mathbb{R}$ be such that the lines $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O=(0,0,0)$, and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$.
Match each entry in List-I to the correct entry in List-II.
| List-I | List-II |
|---|---|
| (P) $\gamma$ equals | (1) $-\hat{i} - \hat{j} + \hat{k}$ |
| (Q) A possible choice for $\hat{n}$ is | (2) $\sqrt{\frac{3}{2}}$ |
| (R) $\overrightarrow{OR_1}$ equals | (3) $1$ |
| (S) A possible value of $\overrightarrow{OR_1} \cdot \hat{n}$ is | (4) $\frac{1}{\sqrt{6}} \hat{i} - \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}$ |
| (5) $\sqrt{\frac{2}{3}}$ |
The correct option is :
Let $\mathbb{R}^3$ denote the three-dimensional space. Take two points $P=(1,2,3)$ and $Q=(4,2,7)$. Let $\operatorname{dist}(X, Y)$ denote the distance between two points $X$ and $Y$ in $\mathbb{R}^3$. Let
$ \begin{gathered} S=\left\{X \in \mathbb{R}^3:(\operatorname{dist}(X, P))^2-(\operatorname{dist}(X, Q))^2=50\right\} \text { and } \\ T=\left\{Y \in \mathbb{R}^3:(\operatorname{dist}(Y, Q))^2-(\operatorname{dist}(Y, P))^2=50\right\} . \end{gathered} $
Then which of the following statements is (are) TRUE?
$-x+2 y-9 z=7$
$-x+3 y+7 z=9$
$-2 x+y+5 z=8$
$-3 x+y+13 z=\lambda$
has a unique solution $x=\alpha, y=\beta, z=\gamma$. Then the distance of the point
$(\alpha, \beta, \gamma)$ from the plane $2 x-2 y+z=\lambda$ is :
the lines $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is 13. Then $8\left|\sum\limits_{\lambda \in S} \lambda\right|$ is equal to :
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 :
the line $\mathrm{L}: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the intercept of $\mathrm{P}$
on the $y$-axis is 1 , then the distance between $\mathrm{P}$ and $\mathrm{L}$ is :
$\left( {\matrix{ \alpha & \beta & \gamma \cr } } \right)\left( {\matrix{ 2 & {10} & 8 \cr 9 & 3 & 8 \cr 8 & 4 & 8 \cr } } \right) = \left( {\matrix{ 0 & 0 & 0 \cr } } \right)$
lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to :
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 :