The line $y=x+1$ meets the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter then ${\left(3r\right)}^2$ is equal to

Plotting the diagram of given data we have,

Let $P$ be the point $\left(-1+\frac{p}{\sqrt{2}},\frac{p}{\sqrt{2}}\right)$

Since it lies on the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$, we get 

$\frac{{\left(-1+\frac{p}{\sqrt{2}}\right)}^2}{4}+\frac{{\left(\frac{p}{\sqrt{2}}\right)}^2}{2}=1 \Rightarrow \frac{3p^2}{8}-\frac{p}{2\sqrt{2}}-\frac{3}{4}=0$

Here ${\left(p_1-p_2\right)}^2={\left(\frac{2\sqrt{2}}{3}\right)}^2-4\left(-2\right)=\frac{8}{9}+8=\frac{80}{9}$

We know, ${\left(p_1-p_2\right)}^2=P{Q}^2$

$\Rightarrow P{Q}^2=\frac{80}{9}$

Now $r=\frac{PQ}{2}\Rightarrow 9r^2=9\frac{{\left(PQ\right)}^2}{4}=\frac{9}{4}\times \frac{80}{9}=20$