The distance of the plane $x+2y-z=2$ from the
point (2, -1, 3) measured in the direction with
direction ratios 2, 2, 1 is

$d.r\text{'}s$ are 2, 2, 1. $d.c\text{'}s$ are $\frac{2}{3},\frac{2}{3},\frac{1}{3}$

The line is $\frac{x-2}{2/3}=\frac{y+1}{2/3}=\frac{z-3}{1/3}=r\text{,}$ where $r$is the distance.

$\begin{array}{l}\therefore x=2+\frac{2r}{3},y=-1+\frac{2r}{3},z=3+\frac{r}{3}\\ x+2y-z=2\\ \Rightarrow 2+\frac{2r}{3}-2+\frac{4r}{3}-3-\frac{r}{3}=2\Rightarrow \frac{5r}{3}=5\Rightarrow r=3.\end{array}$