Equation of parabola is $y^{2}=2 x$, so vertex lies at origin So, co-ordinates of vertex are $\mathrm{A}(0,0)$. Let $\left(x_{1}, y_{1}\right)$ be the co-ordinates of the point $Q$
$\therefore \quad y_{1}^{2}=2 x_{1}$ $. .(\mathrm{i})$
and slope of $\mathrm{PQ}=\frac{\mathrm{y}_{1}-2}{\mathrm{x}_{1}-2}$
[co-ordinates of $\mathrm{P}$ is $(2,2)$ as given]
Also, slope of $\mathrm{AP}=\frac{2-0}{2-0}=1$
Since, $\mathrm{PQ}$ and $\mathrm{AP}$ are perpendicular to each other, hence, slope of AP $\times$ Slope of $P Q=-1$
So, $1 \times\left(\frac{y_{1}-2}{x_{1}-2}\right)=-1$
$\Rightarrow \quad y_{1}-2=-x_{1}+2$
$\Rightarrow x_{1}+y_{1}=4 \Rightarrow x_{1}=4-y_{1}$
Putting value of $\mathrm{x}$, in equation (i)
$y_{1}^{2}=8-2 y_{1}$ or $y_{1}^{2}+2 y_{1}-8=0$
$\Rightarrow \quad y_{1}=-4$ and 2
Hence, co-ordinates of point $\mathrm{Q}$ are $(8,-4)$.
So, required length $P Q=\sqrt{(8-2)^{2}+(-4-2)^{2}}$
$=\sqrt{36+36}=\sqrt{72}=6 \sqrt{2}$