Let y² = 2x the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP) $\left(\mathrm{SQ}\right)=\frac{147}{4}$. Let C be the circle described taking PQ as a diameter. If the equation of a circle C is 64x² + 64y² – $\mathrm{\alpha }$x – 64$\sqrt{3}$y = $\mathrm{\beta }$, then $\mathrm{\beta }$ – $\mathrm{\alpha }$ is equal to _________.

${\mathrm{y}}^2=12\mathrm{x} \mathrm{a}=3 \mathrm{SP}\times \mathrm{SQ}=\frac{147}{4}$
Let P(3t², 6t) and t₁t₂ = –1
(ends of focal chord)
$\text{ So, }\mathrm{Q}\left(\frac{3}{{\mathrm{t}}^2},\frac{-6}{\mathrm{t}}\right)$
S(3, 0)
SP × SQ = PM₁ × QM₂
(dist. from directrix)
$\begin{array}{l}=\left(3+3{\mathrm{t}}^2\right)\left(3+\frac{3}{{\mathrm{t}}^2}\right)=\frac{147}{4}\\ \Rightarrow \frac{{\left(1+{\mathrm{t}}^2\right)}^2}{{\mathrm{t}}^2}=\frac{49}{12}\\ {\mathrm{t}}^2=\frac{3}{4},\frac{4}{3}\\ \mathrm{t}=\pm \frac{\sqrt{3}}{2},\pm \frac{2}{\sqrt{3}}\end{array}$
considering $\mathrm{t}=\frac{-\sqrt{3}}{2}$
$\mathrm{P}\left(\frac{9}{4},-3\sqrt{3}\right)\text{ and }\mathrm{Q}(4,4\sqrt{3})$
Hence, diametric circle:
$\begin{array}{l}(\mathrm{x}-4)\left(\mathrm{x}-\frac{9}{4}\right)+(\mathrm{y}+3\sqrt{3})(\mathrm{y}-4\sqrt{3})=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-\frac{25}{4}\mathrm{x}-\sqrt{3}\mathrm{y}-27=0\\ \Rightarrow \mathrm{\alpha }=400,\mathrm{\beta }=1728\\ \mathrm{\beta }-\mathrm{\alpha }=1328\end{array}$