On plotting the diagram of given value we get,

Now given parabola is $y^2=2x-3\Rightarrow y^2=2\left(x-\frac{3}{2}\right)$

Let a point on parabola be $P=\left(\left(\frac{3}{2}+\frac{t^2}{2}\right),t\right)$

Equation of tangent at $P$ will be $yt=x+\frac{3}{2}+\frac{t^2}{2}-3$

$\because R\left(0,1\right)$ lies on this tangent 

Then $t=\frac{t^2}{2}-\frac{3}{2}$

i.e. $t=3$ or $-1$.

So $P\equiv \left(6,3\right) \& Q\equiv \left(2,-1\right)$

Now triangle $PQR$ is right angled at $Q$, so $\left(2,-1\right)$ is the orthocentre of the triangle.