Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

The equation of line passing through $(1,-2,3)$ and parallel to the given line is $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}$
Suppose it meets the plane $x-y+z=5$ at the point $Q$ given by $(2t+1,3t-2,-6t+3)$
This lies on the plane $x-y+z=5$
$\therefore 2t+1-(3t-2)+(-6t+3)=5$
On solving, we have $7t=1,\Rightarrow t=\frac{1}{7}$
So, the coordinates of $Q$ are $(\frac{9}{7},\frac{-11}{7},\frac{15}{7})$
Hence required distance $PQ=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$