If the tangents drawn at the points P and Q on the parabola $$y^2 = 2x - 3$$ intersect at the point $$R(0, 1)$$, then the orthocentre of the triangle PQR is

First rewrite the parabola as $$y^2=2\left(x-\frac32\right)$$ or $$y^2=4a(x-h)$$, where $$a=\frac12$$ and $$h=\frac32$$.

A point on the parabola corresponding to parameter $$t$$ is $$P\left(\frac32+\frac{t^2}{2},t\right)$$, and the tangent at parameter $$t$$ is $$ty=x-\frac32+\frac{t^2}{2}$$.

Since the tangents at $$P(t_1)$$ and $$Q(t_2)$$ intersect at $$R(0,1)$$, the point $$R$$ must satisfy both tangent equations. Thus $$t=-\frac32+\frac{t^2}{2}$$.

Multiplying by $$2$$, we get $$t^2-2t-3=0$$, which gives $$(t-3)(t+1)=0$$. Hence $$t_1=3$$ and $$t_2=-1$$.

Therefore, $$P=(6,3)$$ and $$Q=(2,-1)$$.

Now $$m_{PQ}=\frac{3-(-1)}{6-2}=1$$. Hence the altitude through $$R$$ has slope $$-1$$ and its equation is $$y=-x+1$$.

Also, $$m_{RQ}=\frac{1-(-1)}{0-2}=-1$$. Hence the altitude through $$P$$ has slope $$1$$ and its equation is $$y-3=x-6$$, i.e., $$y=x-3$$.

Intersecting the two altitudes, we get $$x-3=-x+1$$, which gives $$2x=4$$ and hence $$x=2$$. Substituting in $$y=-x+1$$, we obtain $$y=-1$$.

Therefore, the orthocentre of $$\triangle PQR$$ is $$(2,-1)$$.

Final Answer : $$(2,-1)$$