Let the line L: 2 x + y = α pass through the point of the...

Given: ${x}^{2} + {y}^{2} = 3$ and ${x}^{2} = 2y$

$⇒ {y}^{2} + 2y - 3 = 0$

$⇒ \left(y + 3\right)\left(y - 1\right) = 0$

$⇒ y = 1, - 3\left(\text{Rejected as it gives imaginary value of}x\right)$

$⇒ y = 1$

$⇒ x = \sqrt{2}$

$⇒ P ≡ \left(\sqrt{2}, 1\right)$

Now, $P$ lies on the line $\sqrt{2}x + y = \alpha$.

$⇒ \sqrt{2}\left(\sqrt{2}\right) + 1 = \alpha$

$⇒ \alpha = 3$

For circle ${C}_{1}$, ${Q}_{1}$ lies on $y -$axis.

Let ${Q}_{1} ≡ \left(0, \beta\right)$ and ${R}_{1} = 2\sqrt{3}$ (given)

Line $L$ act as tangent so applying the condition of tangency

$⇒ P = r$

$⇒ \left|\dfrac{\beta - 3}{\sqrt{3}}\right| = 2\sqrt{3}$

$⇒ \left|\beta - 3\right| = 6$

$⇒ \beta - 3 = 6, - 6$

$⇒ \beta = 9, - 3$

So, ${Q}_{1} ≡ \left(0, 9\right) \text{ and } {Q}_{2} ≡ \left(0, - 3\right)$

⇒ Ar\left(\DeltaP{Q}_{1}{Q}_{2}\right) = \dfrac{1}{2}\left|$\begin{array}{ll}\sqrt{2} \\ 0 \\ 0\end{array}$$\begin{array}{ll}1 \\ 9 \\ - 3\end{array}$$\begin{array}{ll}1 \\ 1 \\ 1\end{array}$\right|

$⇒ Ar\left(\DeltaP{Q}_{1}{Q}_{2}\right) = \dfrac{1}{2}\left(\sqrt{2}\left(12\right)\right)$

$⇒ Ar\left(\DeltaP{Q}_{1}{Q}_{2}\right) = 6\sqrt{2}$

$⇒ {\left(\DeltaP{Q}_{1}{Q}_{2}\right)}^{2} = 72$