Let $y^2=12 x$ be the parabola and $S$ be its focus. Let $P Q$ be a focal chord of the parabola such that $(S P)(S Q)=\frac{147}{4}$. Let $C$ be the circle described taking $P Q$ as a diameter. If the equation of a circle $C$ is $64 x^2+64 y^2-\alpha x-64 \sqrt{3} y=\beta$, then $\beta-\alpha$ is equal to $\qquad$ .

<p>$$\mathrm{y}^2=12 \mathrm{x} \quad \mathrm{a}=3 \quad \mathrm{SP} \times \mathrm{SQ}=\frac{147}{4}$$</p> <p>Let $\mathrm{P}\left(3 \mathrm{t}^2, 6 \mathrm{t}\right)$ and $\mathrm{t}_1 \mathrm{t}_2=-1$</p> <p>(ends of focal chord)</p> <p>So, $Q\left(\frac{3}{t^2}, \frac{-6}{t}\right)$</p> <p>$S(3,0)$</p> <p>$\mathrm{SP} \times \mathrm{SQ}=\mathrm{PM}_1 \times \mathrm{QM}_2$</p> <p>$$\begin{aligned} & \text { (dist. from directrix) } \\ & =\left(3+3 \mathrm{t}^2\right)\left(3+\frac{3}{\mathrm{t}^2}\right)=\frac{147}{4} \\ & \Rightarrow \frac{\left(1+\mathrm{t}^2\right)^2}{\mathrm{t}^2}=\frac{49}{12} \\ & \mathrm{t}^2=\frac{3}{4}, \frac{4}{3} \\ & \mathrm{t}= \pm \frac{\sqrt{3}}{2}, \pm \frac{2}{\sqrt{3}} \\ & \text { considering } \mathrm{t}=\frac{-\sqrt{3}}{2} \\ & \mathrm{P}\left(\frac{9}{4},-3 \sqrt{3}\right) \text { and } \mathrm{Q}(4,4 \sqrt{3}) \end{aligned}$$</p> <p>$$\begin{aligned} &\text { Hence, diametric circle: }\\ &\begin{aligned} & (x-4)\left(x-\frac{9}{4}\right)+(y+3 \sqrt{3})(y-4 \sqrt{3})=0 \\ & \Rightarrow x^2+y^2-\frac{25}{4} x-\sqrt{3} y-27=0 \\ & \Rightarrow \alpha=400, \beta=1728 \\ & \beta-\alpha=1328 \end{aligned} \end{aligned}$$</p>