We have,

$y^2=2x$

Tangent at $P\left(2,2\right)$ is

$y\left(2\right)=2\left(\frac{1}{2}\right)\left(x+2\right)$

$\Rightarrow 2y=x+2$

It meets $x-$axis at $y=0$ i.e., $x=-2$

$\therefore Q\equiv \left(-2,0\right)$

Normal at $P$ is

$\Rightarrow y-2=-2\left(x-2\right)$

$\Rightarrow y=6-2x$

$\therefore$  Solving with $y^2=2x$, we get

${\left(6-2x\right)}^2=2x$

$\Rightarrow 4x^2+36-24x=2x$

$\Rightarrow 4x^2-26x+36=0$

$\Rightarrow 2x^2-13x+18=0$

$\Rightarrow x=\frac{13\pm \sqrt{169-144}}{4}=\frac{13\pm 5}{4}$

$\Rightarrow x=\frac{9}{2}, \frac{1}{2}$

$\therefore R\equiv \left(\frac{9}{2},-3\right)$

$\therefore Ar\left(\Delta PQR\right)=\frac{1}{2}\left|\begin{array}{ccc}2& 2& 1\\ -2& 0& 1\\ \frac{9}{2}& -3& 1\end{array}\right|$

$\Rightarrow Ar\left(\Delta PQR\right)=\frac{1}{2}\left[2·3-2\left(-2-\frac{9}{2}\right)+6\right]$

$\Rightarrow Ar\left(\Delta PQR\right)=\frac{1}{2}\left[12+13\right]$
$\Rightarrow Ar\left(\Delta PQR\right)=\frac{25}{2}$ sq.units