$\text{ The circle }{C}_1:x^2+y^2=3\text{ with centre at }O,\text{ intersects the parabola }{x}^2=2y\text{ at the point }$ $\mathrm{P}\text{ in the first quadrant. Let the tangent to the circle }{C}_1\text{ at }\mathrm{P}\text{ touches other two circles }{\mathrm{C}}_2$$\text{ and }{C}_3\text{ at }{R}_2\text{ and }{R}_3\text{ respectively. Suppose }{C}_2\text{ and }{C}_3\text{ have equal radii }2\sqrt{3}\text{ and }$$\text{ centers }{Q}_2\text{ and }{Q}_3\text{ respectively. If }{Q}_2\text{ and }{Q}_3\text{ lie on the y-axis, then }$

$x^2+y^2=3,and x^2=2y \text{ intersect at }p(\sqrt{2},1) which lies in the first quadrant$ 

$Equation of \tan gent at P to the above circle is \sqrt{2}x+y=3$

$\text{ Any circle with center on y-axis, radius }2\sqrt{3}\text{ is }{x}^2+(y-f{)}^2=12$

$The line \sqrt{2}x+y=3 touches the above circle$

$\begin{array}{r}\frac{|0+f-3|}{\sqrt{3}}=2\sqrt{3}\Rightarrow f-3=\pm 6\\ f=9,-3\end{array}$

$$\begin{gathered} The equations of C_2,C_{3 }are as below \\ {x}^2+{\left(y-9\right)}^2=12, and x^2+{\left(y+3\right)}^2=12 \end{gathered}$$

$Q_1(0,9)Q_2\left(0,-3\right)\Rightarrow Q_2{Q}_3=12$

$△P{Q}_2{Q}_3=\frac{1}{2}\left|9\sqrt{2}+3\sqrt{2}\right|=\frac{1}{2}\left(12\sqrt{2}\right)=6\sqrt{2}$

$\begin{array}{l}{R}_1\left(h_{1}k\right) is point of contact, it is foot of the perpendicular of center on the \tan gent\\ \frac{h-0}{\sqrt{2}}=\frac{k-9}{1}=-\left(\frac{6}{3}\right)=-2\\ R_1\left(-2\sqrt{2},7\right)\\ R_2\left(h,k\right) is point of contact, it is foot of the perpendicular of center on the \tan gent\\ \frac{h-0}{\sqrt{2}}=\frac{k+3}{1}=2\\ R_2\left(2\sqrt{2},-1\right)\\ R_2{R}_3=4\sqrt{6}\end{array}$

$\text{ Area }\Delta O{R}_2{R}_3\text{ is }6\sqrt{2}$