Let the line $\mathrm{L}: \sqrt{2} x+y=\alpha$ pass through the point of the intersection $\mathrm{P}$ (in the first quadrant) of the circle $x^2+y^2=3$ and the parabola $x^2=2 y$. Let the line $\mathrm{L}$ touch two circles $\mathrm{C}_1$ and $\mathrm{C}_2$ of equal radius $2 \sqrt{3}$. If the centres $Q_1$ and $Q_2$ of the circles $C_1$ and $C_2$ lie on the $y$-axis, then the square of the area of the triangle $\mathrm{PQ}_1 \mathrm{Q}_2$ is equal to ___________.

<p>$x^2+y^2=3$ and $x^2=2 y$</p> $y^2+2 y-3=0 $ <br/><br/>$\Rightarrow(y+3)(y-1)=0$ <br/><br/>$y=-3$ (Rejected) or $y=1$ <br/><br/>For $\mathrm{y}=1, \mathrm{x}=\sqrt{2} \Rightarrow P(\sqrt{2}, 1)$ <br/><br/>$p$ lies on the line <br/><br/>$$ \begin{aligned} & \sqrt{2} x+y=\alpha \\\\ & \sqrt{2}(\sqrt{2})+1=\alpha \\\\ & \alpha=3 \end{aligned} $$ <br/><br/>For circle $\mathrm{C}_1$ <br/><br/>$\mathrm{Q}_1$ lies on $\mathrm{y}$ axis <br/><br/>Let $\mathrm{Q}_1(0, \alpha)$ coordinates <br/><br/>$\mathrm{R}_1=2 \sqrt{3}$ (Given <br/><br/>Line $\mathrm{L}$ act as tangent <br/><br/>Apply P $=r$ (condition of tangency) <br/><br/>$\begin{aligned} & \Rightarrow\left|\frac{\alpha-3}{\sqrt{3}}\right|=2 \sqrt{3} \\\\ & \Rightarrow|\alpha-3|=6\end{aligned}$ <br/><br/>$\therefore$ $\alpha-3=6$ <br/><br/>$\Rightarrow \alpha=9$ <br/><br/>$\begin{gathered}\text { or } \alpha-3=-6 \\\\ \Rightarrow \alpha=-3\end{gathered}$ <br/><br/>$\begin{aligned} & \triangle P Q_1 Q_2=\frac{1}{2}\left|\begin{array}{ccc}\sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1\end{array}\right| \\\\ & =\frac{1}{2}(\sqrt{2}(12))=6 \sqrt{2} \\\\ & \left(\triangle P Q_1 Q_2\right)^2=72\end{aligned}$