Since direction cosines of PQ are equal and positive.
$\therefore \quad$ The d.r.s. of PQ are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$
$\therefore \quad$ The equation of the line $\mathrm{PQ}$ is
$$
\begin{aligned}
& \frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{\mathrm{z}-2}{\frac{1}{\sqrt{3}}} \\
& \Rightarrow x-2=y+1=\mathrm{z}-2=\mathrm{k}, \text { say }
\end{aligned}
$$
$\therefore \quad$ Co-ordinates of the point $\mathrm{Q}$ are
$$
(\mathrm{k}+2, \mathrm{k}-1, \mathrm{k}+2)
$$

The point $\mathrm{Q}$ lies on the plane $2 x+y+z=9$
$$
\begin{array}{ll}
\therefore & 2(\mathrm{k}+2)+\mathrm{k}-1+\mathrm{k}+2=9 \\
& \Rightarrow 4 \mathrm{k}+5=9 \quad \Rightarrow \mathrm{k}=1 \\
\therefore & \mathrm{Q} \equiv(3,0,3) \\
\therefore & \mathrm{PQ}=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\
& =\sqrt{1+1+1}=\sqrt{3}
\end{array}
$$