Let chord PQ of length $$3\sqrt{13}$$ of the parabola $$y^2 = 12x$$ be such that the ordinates of points P and Q are in the ratio 1:2. If the chord PQ subtends an angle $$\alpha$$ at the focus of the parabola, then $$\sin \alpha$$ is equal to:

$$4a = 12 \implies a = 3$$

$$P = (3t_1^2, 6t_1) \quad \text{and} \quad Q = (3t_2^2, 6t_2)$$

$$\frac{6t_1}{6t_2} = \frac{1}{2} \implies t_2 = 2t_1$$

$$Q = (3(2t_1)^2, 6(2t_1)) = (12t_1^2, 12t_1)$$

$$PQ^2 = (12t_1^2 - 3t_1^2)^2 + (12t_1 - 6t_1)^2$$

$$(3\sqrt{13})^2 = (9t_1^2)^2 + (6t_1)^2$$

$$(9t_1^2 + 13)(t_1^2 - 1) = 0$$

$$t_1^2 = 1 \implies t_1 = 1 \quad (\text{taking positive parameter for simplicity})$$

Thus, our parameters are $$t_1 = 1$$ and $$t_2 = 2$$

Using the property $$F(x_1, y_1) = a(1 + t^2)$$:

$$FP = 3(1 + t_1^2) = 3(1 + 1) = 6$$

$$FQ = 3(1 + t_2^2) = 3(1 + 2^2) = 15$$

$$\cos\alpha = \frac{FP^2 + FQ^2 - PQ^2}{2 \cdot FP \cdot FQ}$$ (Cosine rule)

$$\cos\alpha = \frac{6^2 + 15^2 - (3\sqrt{13})^2}{2 \cdot 6 \cdot 15}$$

$$\cos\alpha = \frac{36 + 255 - 117}{180} = \frac{144}{180} = \frac{4}{5}$$

$$\sin\alpha = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$$