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

We need to find the orthocentre of triangle PQR, where P and Q lie on the parabola $$y^2 = 2x - 3 and the tangents at P and Q intersect at R(0, 1)$$.

Since rewriting the equation as $$y^2 = 2\left(x - \frac{3}{2}\right) shows a parabola with vertex at \left(\frac{3}{2}, 0\right) and parameter 4a = 2 so that a = \frac{1}{2}, the parametric form can be taken as x = \frac{3}{2} + \frac{t^2}{2}, \quad y = t$$.

For the parabola $$y^2 = 2\left(x - \frac{3}{2}\right), the tangent at parameter t is given by ty = \left(x - \frac{3}{2}\right) + \frac{t^2}{2}\,.$$

Substituting $$R(0,1) into this tangent equation yields t \cdot 1 = \left(0 - \frac{3}{2}\right) + \frac{t^2}{2}\,,\; or t = -\frac{3}{2} + \frac{t^2}{2}\,,\; which simplifies to t^2 - 2t - 3 = 0\,. Factoring gives (t - 3)(t + 1) = 0\,,\; so t = 3\text{ or }t = -1\,. Therefore, for t = 3 we get P = \left(\frac{3}{2} + \frac{9}{2}, 3\right) = (6, 3)\,,\, and for t = -1 we get Q = \left(\frac{3}{2} + \frac{1}{2}, -1\right) = (2, -1)\,. With vertices P(6,3)\,,\;Q(2,-1)\,,\;R(0,1)\, the slope of PQ is \frac{3 - (-1)}{6 - 2} = 1, so the altitude from R has slope -1 and its equation is y - 1 = -1(x - 0)\;\Rightarrow\;y = -x + 1\quad\text{(i)}\,.$$

Next, the slope of $$PR is \frac{3 - 1}{6 - 0} = \frac{1}{3}\,,\, so the altitude from Q has slope -3 and its equation is y - (-1) = -3(x - 2)\;\Rightarrow\;y = -3x + 5\quad\text{(ii)}\,. Equating (i) and (ii) gives -x + 1 = -3x + 5\,,\; hence 2x = 4\,,\;x = 2\,,\, and substituting back into y = -x + 1 yields y = -2 + 1 = -1\,. Therefore, the orthocentre of triangle PQR is (2, -1)\,, which corresponds to Option B.