Let C be the centre of the circle $$x^2 + y^2 - x + 2y = \frac{11}{4}$$ and P be a point on the circle. A line passes through the point C, makes an angle of $$\frac{\pi}{4}$$ with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit$$^2$$) is

We have the circle $$x^2 + y^2 - x + 2y = \frac{11}{4}$$ and need to find the area of triangle PQR.

First, rewrite the equation by completing the squares: $$\left(x - \frac{1}{2}\right)^2 + (y + 1)^2 = \frac{11}{4} + \frac{1}{4} + 1 = 4$$, which shows that the centre is $$C = \left(\frac{1}{2}, -1\right)$$ and the radius is $$r = 2$$.

Let P be any point on the circle. A line passing through the centre C makes an angle of $$\frac{\pi}{4}$$ with the line CP and intersects the circle again at Q and R. Since this line goes through the centre, QR is a diameter of the circle, so $$|QR| = 2r = 4$$.

To determine the height of triangle PQR from P to QR, observe that the angle between CP and QR is $$\frac{\pi}{4}$$ and that $$|CP| = r = 2$$ because P lies on the circle. The perpendicular distance from P to the line QR is therefore $$h = |CP| \sin\frac{\pi}{4} = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$$.

Using the base QR and the corresponding height h, the area of triangle PQR is calculated as $$\text{Area} = \frac{1}{2} \times |QR| \times h = \frac{1}{2} \times 4 \times \sqrt{2} = 2\sqrt{2}$$.

Therefore, the correct answer is Option B: $$2\sqrt{2}$$.