Let the tangent to the parabola $\mathrm{y}^{2}=12 \mathrm{x}$ at the point $(3, \alpha)$ be perpendicular to the line $2 x+2 y=3$. Then the square of distance of the point $(6,-4)$ from the normal to the hyperbola $\alpha^{2} x^{2}-9 y^{2}=9 \alpha^{2}$ at its point $(\alpha-1, \alpha+2)$ is equal to _________.

$\because \mathrm{P}(3, \alpha)$ lies on $\mathrm{y}^2=12 \mathrm{x}$ <br/><br/>$\Rightarrow \alpha= \pm 6$ <br/><br/>$$ \text { But, }\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(3, \alpha)}=\frac{6}{\alpha}=1 \Rightarrow \alpha=6(\alpha=-6 \text { reject }) $$ <br/><br/>Now, hyperbola $\frac{x^2}{9}-\frac{y^2}{36}=1$, normal at <br/><br/>$$ \begin{aligned} & \mathrm{Q}(\alpha-1, \alpha+2) \text { is } \frac{9 \mathrm{x}}{5}+\frac{36 y}{8}=45 \\\\ & \Rightarrow 2 x+5 y-50=0 \end{aligned} $$ <br/><br/>Now, distance of $(6,-4)$ from $2 x+5 y-50=0$ is equal to <br/><br/>$$ \begin{aligned} & \left|\frac{2(6)-5(4)-50}{\sqrt{2^2+5^2}}\right|=\frac{58}{\sqrt{29}} \\\\ & \Rightarrow \text { Squareof distance }=116 \end{aligned} $$