Angle Subtended by Parabola Chord at Focus

The equation of the given parabola is $y^2 = 12x$ Comparing with $y^2 = 4ax$, we have $4a = 12 \implies a = 3$. Thus, the focus of the parabola is $S(3, 0)$.

Let the ordinates of points $P$ and $Q$ be $y_1$ and $y_2$ respectively. We are given that $y_1 : y_2 = 1 : 2$. Let $y_1 = k$ and $y_2 = 2k$.

Since both points lie on the parabola $y^2 = 12x$, their $x$-coordinates are given by $x = \frac{y^2}{12}$: $x_1 = \frac{k^2}{12}, \quad x_2 = \frac{(2k)^2}{12} = \frac{k^2}{3}$

So, the coordinates of $P$ and $Q$ are: $P\left(\frac{k^2}{12}, k\right) \quad \text{and} \quad Q\left(\frac{k^2}{3}, 2k\right)$

The length of the chord $PQ$ is given as $3\sqrt{13}$. Using the distance formula: $PQ^2 = \left(\frac{k^2}{3} - \frac{k^2}{12}\right)^2 + (2k - k)^2 = (3\sqrt{13})^2$ $\left(\frac{k^2}{4}\right)^2 + k^2 = 117$ $\frac{k^4}{16} + k^2 - 117 = 0$ $k^4 + 16k^2 - 1872 = 0$

Factoring the quadratic equation in $k^2$: $(k^2 + 52)(k^2 - 36) = 0$

Since $k^2 > 0$, we get: $k^2 = 36 \implies k = \pm 6$

Taking $k = 6$ (the case $k = -6$ yields symmetric points with the same geometric result): $P = (3, 6) \quad \text{and} \quad Q = (12, 12)$

Now, consider the vectors $\vec{SP}$ and $\vec{SQ}$ from the focus $S(3, 0)$: $\vec{SP} = (3 - 3)\hat{i} + (6 - 0)\hat{j} = 6\hat{j}$ $\vec{SQ} = (12 - 3)\hat{i} + (12 - 0)\hat{j} = 9\hat{i} + 12\hat{j}$

The magnitudes of these vectors are: $|\vec{SP}| = 6$ $|\vec{SQ}| = \sqrt{9^2 + 12^2} = 15$

The dot product of $\vec{SP}$ and $\vec{SQ}$ is: $\vec{SP} \cdot \vec{SQ} = (0)(9) + (6)(12) = 72$

If $\alpha$ is the angle subtended by $PQ$ at the focus $S$, then: $\cos \alpha = \frac{\vec{SP} \cdot \vec{SQ}}{|\vec{SP}| |\vec{SQ}|} = \frac{72}{6 \times 15} = \frac{72}{90} = \frac{4}{5}$

Since $\sin \alpha = \sqrt{1 - \cos^2 \alpha}$: $\sin \alpha = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25}} = \frac{3}{5}$

Thus, $\sin \alpha$ is equal to $\frac{3}{5}$.