Given equation of parabola is

$y^2=3x$

$\Rightarrow 2y\frac{dy}{dx}=3$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\mathrm{slope} \mathrm{of} \mathrm{tangent} \mathrm{at} P}=\frac{3}{2y}$

Tangent at $P\left(x_1,y_1\right)$ is parallel to the line $x+2y=1$, so slope of tangent is

$m=-\frac{1}{2}$

$\Rightarrow \frac{3}{2y_1}=-\frac{1}{2}$

$\Rightarrow y_1=-3$

So, coordinates of $P\equiv \left(3,-3\right)$.

Given equation of ellipse is

$\frac{x^2}{4}+\frac{y^2}{1}=1$

$\Rightarrow x^2+4y^2=4$

$\Rightarrow 2x+8y\left(\frac{dy}{dx}\right)=0$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\mathrm{slope} \mathrm{of} \mathrm{tangent}}=-\frac{x}{4y}$

Also, tangent to ellipse re perpendicular to line $x-y=2$, so slope of tangent

$m=-1$

$\Rightarrow -\frac{x}{4y}=-1$

$\Rightarrow x=4y$

Putting in equation of ellipse, we get

$\frac{16y^2}{4}+\frac{y^2}{1}=1$

$\Rightarrow y=\pm \frac{1}{\sqrt{5}}$

So,

$Q\left(\frac{4}{\sqrt{3}},\frac{1}{\sqrt{5}}\right),R\left(-\frac{4}{\sqrt{5}},\frac{-1}{\sqrt{5}}\right)$

Area of $∆PQR$ is

$=\frac{1}{2}\left|\begin{array}{ccc}3& -3& 1\\ \frac{4}{\sqrt{5}}& \frac{1}{\sqrt{5}}& 1\\ -\frac{4}{\sqrt{5}}& -\frac{1}{\sqrt{5}}& 1\end{array}\right|$

$=\frac{1}{2}\left[3\left(\frac{2}{\sqrt{5}}\right)+3\left(\frac{8}{\sqrt{5}}\right)+0\right]$

$=\frac{30}{2\sqrt{5}}=3\sqrt{5}$ sq. units