1. Find the intersection point $P$ in the first quadrant

We are given: $x^2+y^2=3$ and $x^2=2y \implies y=\frac{x^2}{2}$

Substitute into the circle: $x^2+\left(\frac{x^2}{2}\right)^2=3$ $x^2+\frac{x^4}{4}=3$ $4x^2+x^4=12$

Let $t=x^2$. Then: $t^2+4t-12=0$ $(t+6)(t-2)=0$ Since $x^2\ge 0$, we get: $x^2=2 \implies x=\sqrt{2}$ (in the first quadrant)

Then: $y=\frac{x^2}{2}=1$ So, $P=(\sqrt{2},1)$

2. Find the line $L$ passing through $P$

The line is: $\sqrt{2}x+y=\alpha$ Since it passes through $P=(\sqrt{2},1)$, $\alpha=\sqrt{2}(\sqrt{2})+1=2+1=3$ Hence, $L:\ \sqrt{2}x+y=3$

3. Find centres on the $y$-axis of circles of radius $2\sqrt{3}$ tangent to $L$

Let the centre be $(0,k)$ since it lies on the $y$-axis.

Distance from $(0,k)$ to the line $\sqrt{2}x+y-3=0$ must equal the radius $2\sqrt{3}$.

Using point-to-line distance: $\frac{|\sqrt{2}(0)+k-3|}{\sqrt{(\sqrt{2})^2+1^2}}=2\sqrt{3}$ $\frac{|k-3|}{\sqrt{3}}=2\sqrt{3}$ $|k-3|=6$ So, $k=9 \quad \text{or} \quad k=-3$

Therefore, $Q_1=(0,9),\qquad Q_2=(0,-3)$

4. Find area of triangle $PQ_1Q_2$

Since $Q_1,Q_2$ lie on the $y$-axis, the segment $Q_1Q_2$ is vertical.

Its length is: $Q_1Q_2=|9-(-3)|=12$

The perpendicular distance of $P=(\sqrt{2},1)$ from the $y$-axis is its $x$-coordinate: $\sqrt{2}$

Hence area of triangle: $\Delta=\frac{1}{2}\times 12\times \sqrt{2}=6\sqrt{2}$

Therefore, the square of the area is: $\Delta^2=(6\sqrt{2})^2=72$

5. Compare with stored answer

Derived answer: $72$

Stored correct answer: $72$

They agree.