Question:
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 :
Correct Option: , 2
Solution:
equation of line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ passes
through $(1,-2,3)$ is
$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$
$x=2 r+1$
$y=3 r-2$
$z=-6 r+3$
So $2 r+1-3 r+2-6 r+3=5$
$\Rightarrow$ $-7 r+1=0$
$r=\frac{1}{7}$
$x=\frac{9}{7}, y=\frac{-11}{7},, z=\frac{15}{7}$
Distance is $=\sqrt{\left(\frac{9}{7}-1\right)^{2}+\left(2-\frac{11}{7}\right)^{2}+\left(3-\frac{15}{7}\right)^{2}}$
$=\sqrt{\left(\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}+\left(\frac{6}{7}\right)^{2}}$
$=\frac{1}{7} \sqrt{4+9+36}$
$=\frac{1}{7} \sqrt{49}=1$