Let the line $$L: \sqrt{2}x + y = \alpha$$ pass through the point of the intersection $$P$$ (in the first quadrant) of the circle $$x^2 + y^2 = 3$$ and the parabola $$x^2 = 2y$$. Let the line $$L$$ touch two circles $$C_1$$ and $$C_2$$ 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 $$PQ_1Q_2$$ is equal to:

We need to find the square of the area of triangle $$PQ_1Q_2$$.

First, we find the point P in the first quadrant where the circle $$x^2 + y^2 = 3$$ intersects the parabola $$x^2 = 2y$$. Substituting $$x^2 = 2y$$ into the circle gives $$2y + y^2 = 3$$, so $$y^2 + 2y - 3 = 0$$ and $$(y+3)(y-1) = 0$$. Since P is in the first quadrant, $$y = 1$$ and $$x^2 = 2$$, hence $$x = \sqrt{2}$$, which gives $$P = (\sqrt{2},1)$$.

The line $$L: \sqrt{2}x + y = \alpha$$ passes through $$P(\sqrt{2},1)$$, so $$\sqrt{2}\cdot\sqrt{2} + 1 = 2 + 1 = 3$$, giving $$\alpha = 3$$. Therefore the equation of the line is $$\sqrt{2}x + y = 3$$.

The circles $$C_1$$ and $$C_2$$ each have radius $$2\sqrt{3}$$ and centres on the y-axis, denoted $$Q_1 = (0,k_1)$$ and $$Q_2 = (0,k_2)$$. Since line $$L$$ is tangent to each circle, the distance from $$(0,k)$$ to the line $$\sqrt{2}x + y - 3 = 0$$ must equal $$2\sqrt{3}$$. This distance is $$d = \frac{|0 + k - 3|}{\sqrt{2 + 1}} = \frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3}$$, which implies $$|k - 3| = 6$$. Hence $$k = 9$$ or $$k = -3$$, giving $$Q_1 = (0,9)$$ and $$Q_2 = (0,-3)$$.

With $$P = (\sqrt{2},1)$$, $$Q_1 = (0,9)$$ and $$Q_2 = (0,-3)$$, the base $$Q_1Q_2$$ has length $$|9 - (-3)| = 12$$ and the height from P to the y-axis is $$|\sqrt{2}| = \sqrt{2}$$. Therefore the area of triangle $$PQ_1Q_2$$ is $$\frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$$.

Finally, squaring this area gives $$(6\sqrt{2})^2 = 72$$. Thus, the square of the area is 72.