The distance of a point A−2,3,1 from the line  MN→ throughM−3,5,2 , which makes equal angles with the coordinate axes is

# The distance of a point $A\left(-2,3,1\right)$ from the line  $\stackrel{\to }{MN}$ through$M\left(-3,5,2\right)$ , which makes equal angles with the coordinate axes is

1. A

$\frac{\sqrt{16}}{\sqrt{3}}$

2. B

$\frac{\sqrt{14}}{\sqrt{3}}$

3. C

$\frac{\sqrt{2}}{\sqrt{3}}$

4. D

$\frac{\sqrt{5}}{\sqrt{3}}$

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### Solution:

The equation of line passing through $M\left(-3,5,2\right)$ and making equal angles with coordinate axes is $\frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}$

General point on the above line is$P\left(k-3,k+5,k+2\right)$

If the perpendicular distance from  $A\left(-2,3,1\right)$ to the above line is $AP$  then  $AP$is perpendicular to the line  $\frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}$

Hence,

$\begin{array}{c}\left(k-1\right)\cdot 1+\left(k+2\right)\cdot 1+\left(k+1\right)\cdot 1=0\\ 3k+2=0\\ k=-\frac{2}{3}\end{array}$

Hence the point  $P\left(k-3,k+5,k+2\right)$ is $P\left(-\frac{11}{3},-\frac{13}{5},-\frac{4}{3}\right)$

The required distance is $AP$,  $AP=\sqrt{\frac{25}{9}+\frac{16}{9}+\frac{1}{9}}=\overline{)\sqrt{\frac{14}{3}}}$

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