A line passes through the point $P(2,-1,2)$ and makes equal angles with positive axes. It meets the plane $2 x+y+z=9$ 

at point $Q$ The $PQ$ is 

The d c's of the line line are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.

Then line is $\frac{x-2}{1/\sqrt{3}}=\frac{y+1}{1/\sqrt{3}}=\frac{z-2}{1/\sqrt{3}}=r$, distance

$\therefore x=2+\frac{r}{\sqrt{3}},y=-1+\frac{r}{\sqrt{3}},z=2+\frac{r}{\sqrt{3}}$

$2 x+y+z=9$

$\Rightarrow 4+\frac{2r}{\sqrt{3}}-1+\frac{r}{\sqrt{3}}+2+\frac{r}{\sqrt{3}}=9$

$\Rightarrow \frac{4r}{\sqrt{3}}=4\Rightarrow r=\sqrt{3}.$