Question

Distance of the point $(1,–2,3)$ from the plane $x–y+z=5$ measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is

• A. $2$
• B. $3$
• C. $1$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is C $1$


Eqn. PQ is $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$
$\therefore Q=(2r+1,3r-2,-6r+3)$
$\Rightarrow (2r+1)-(3r-2)+(-6r+3)=5\Rightarrow r=\frac{1}{7}$
$\therefore Q=(\frac{9}{7},\frac{-11}{7},\frac{15}{7})$
$\therefore PQ=1$