Given equations of plane are $-x+2y-z=0$ and $3x-5y+2z=0$

Now a vector perpendicular to both the planes is given by $\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 2& -1\\ 3& -5& 2\end{array}\right|$

$=\hat{i}\left(-1\right)-\hat{j}\left(1\right)+\hat{k}\left(-1\right)$

$\overrightarrow{n}=-\hat{i}-\hat{j}-\hat{k}$

For a point on the line of intersection of both the planes

let $z=0$,

so $x=2y \& 8x=5y$

i.e. a point on the line will be $\left(0,0,0\right)$

So equation of line of intersection of both these planes will be $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$

The DR's of $QR$ are $1, 1, 1$

Let the point $T\left(\alpha ,\alpha ,\alpha \right)$ be the foot of perpendicular from $P$ to the line

$\Rightarrow \left(\alpha -1\right)\times 1+\left(\alpha +2\right)\times 1+\left(\alpha -3\right)\times 1=0$

i.e. $\alpha =\frac{2}{3}$

or $T\equiv \left(\frac{2}{3},\frac{2}{3},\frac{2}{3}\right)$

$P{T}^2=\frac{1}{9}+\frac{64}{9}+\frac{49}{9}$

$P{T}^2=\frac{114}{9}$

$PT=\frac{\sqrt{114}}{3}$

If $\theta$ is $\angle TPQ$, then in right anged triangle $TPQ$, 

$\cos \theta =\frac{\sqrt{114}}{3\times 3\sqrt{2}}=\frac{\sqrt{57}}{9}=\frac{\sqrt{19}}{3\sqrt{3}}$

Now $\cos 2\theta =2{\cos }^2\theta -1=\frac{2\times 19}{27}-1=\frac{11}{27}$

and $\sin 2\theta =\sqrt{1-{\left(\frac{11}{27}\right)}^2}$$=\frac{4}{27}\sqrt{38}$

So required area of triangle $PQR$ $=\frac{1}{2}\times \sqrt{18}\sqrt{18}\times \frac{4}{27}\sqrt{38}$

$=\frac{18}{2}\times \frac{4}{27}\sqrt{38}=\frac{4}{3}\sqrt{38}$