The circle C _ 1 : x ^ 2 + y ^ 2 = 3, with centre at O, intersects...

On solving ${x}^{2} + {y}^{2} = 3 and {x}^{2} = 2y$ we get point $P\left(\sqrt{2}, 1\right)<br >Equation$ of tangent at P
\sqrt{2}x + y = 3$<br >Let${Q}_{2}$be (0, k) and radius is$2\sqrt{3}$<br >$∴ \left|\dfrac{\sqrt{2}\left(0\right)+ k - 3}{\sqrt{2 + 1}}\right| = 2\sqrt{3}
$∴ k = 9, - 3 {Q}_{3}$
${Q}_{2}\left(0, 9\right)$ and {Q}_{3} ( 0, - 3 ) { $$\begin{array}{ll}{C}_{2} : {( x - 0 )}^{2} + {( y - 9 )}^{2} = 12 \\ {C}_{3} : {( x - 0 )}^{2} + {( y + 3 )}^{2} = 12\end{array}$$
Hence ${Q}_{2}{Q}_{3} = 12$

${⇒ a}^{2} + 12 = 36$
$⇒ a = 2\sqrt{6}$
${R}_{2}{R}_{3} = 2a = 4\sqrt{6}$
Perpendicular distance of origin O from ${R}_{2}{R}_{3}$ is equal to distance of O from tangent $\sqrt{2}x + y = 3$ which is same as radius of circle ${C}_{1} = \sqrt{3}$
Hence area of $\DeltaO{R}_{2}{R}_{3} = \dfrac{1}{2} \times \left({R}_{2}{R}_{3}\right)\sqrt{3} = \dfrac{1}{2} . 4\sqrt{6} . \sqrt{3} = 6\sqrt{2}$
Perpendicular Distance of P from ${Q}_{2}{Q}_{3} = \sqrt{2}$
$∴$ Area of $\DeltaP{Q}_{2}{Q}_{3} = \dfrac{1}{2} \times$ 12 \times \sqrt{2} = 6\sqrt{2}