Step-by-Step Solution
Key Concept: For points on a focal chord, if their distances from the focus are in H.P., use the focal chord property $\frac{1}{r_1} + \frac{1}{r_2} = \frac{2}{l}$.
Since $PS$, $SQ$ are in H.P., we have $\frac{1}{PS} + \frac{1}{SQ}$ are in A.P. For a focal chord, $\frac{1}{PS} + \frac{1}{SQ} = \frac{2}{l}$ where $l$ is the semi-latus rectum. We are given that $PQ = 6 + 3 = 9 = 4\text{sec}^2\theta$ (where $\tan\theta$ is the slope of $PQ$). From the harmonic progression condition and properties of focal chords, we get $1 + \cot^2\theta = \frac{4}{3}$, leading to $\cot^2\theta = 1$.
Correct Answer: 3