Let C 1 and C 2 be the centres of the circles x 2 + y 2 - 2 x - 2 y - 2 = 0 and x 2 + y 2 - 6 x - 6 y + 14 =…

Equation of given circles are

${\left(x-1\right)}^2+{\left(y-1\right)}^2=4$ and ${\left(x-3\right)}^2+{\left(y-3\right)}^2=4$.

Hence, $C_1\left(1,1\right)$ and $r_1=2$; $C_2\left(3,3\right)$ and $r_2=2$

$\Rightarrow P{C}_1=P{C}_2=2$

Now, by distance formula,

$C_1{C}_2=\sqrt{{\left(3-1\right)}^2+{\left(3-1\right)}^2}=\sqrt{2^2+2^2}=\sqrt{8}$

$\Rightarrow P{C}_1^2+P{C}_2^2=C_1{C}_2^2$

$\Rightarrow \angle C_{1}P{C}_2=\frac{\pi }{2}$ (by converse of pythagoras theorem in $△P{C}_1{C}_2$)

Hence, area of quadrilateral $P{C}_{1}Q{C}_2=2\times$area of $△P{C}_1{C}_2=2\times$area of $△Q{C}_1{C}_2$

$=2\times \frac{1}{2}\times 2\times 2=4$