The distance of the point


The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line

$\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is :


  1. 7

  2. 1

  3. $\frac{1}{7}$

  4. $\frac{7}{5}$

Correct Option: , 2


equation of line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ passes

through $(1,-2,3)$ is


$x=2 r+1$

$y=3 r-2$

$z=-6 r+3$

So $2 r+1-3 r+2-6 r+3=5$

$\Rightarrow$ $-7 r+1=0$


$x=\frac{9}{7}, y=\frac{-11}{7},, z=\frac{15}{7}$

Distance is $=\sqrt{\left(\frac{9}{7}-1\right)^{2}+\left(2-\frac{11}{7}\right)^{2}+\left(3-\frac{15}{7}\right)^{2}}$


$=\frac{1}{7} \sqrt{4+9+36}$

$=\frac{1}{7} \sqrt{49}=1$

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