Let $P\left(a_1, b_1\right)$ and $Q\left(a_2, b_2\right)$ be two distinct points on a circle with center $C(\sqrt{2},\sqrt{3})$. Let $O$ be the origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of the triangle $OCP$ is $\frac{\sqrt{35}}{2}$, then $a_1^2+a_2^2+b_1^2+b_2^2$ is equal to  __________

Given,

$P\left(a_1, b_1\right)$ and $Q\left(a_2, b_2\right)$ be two distinct points on a circle with center $C(\sqrt{2},\sqrt{3})$,

And $O$ be the origin and $OC$ be perpendicular to both $CP$ and $CQ$,

So, $PCQ$ will be a straight line

Now on plotting the diagram we get,

So, $OC=\sqrt{{\left(\sqrt{2}\right)}^2+{\left(\sqrt{3}\right)}^2}=\sqrt{5}$

Now let $CP=CQ=I$
Now by using area of triangle $OCP=\frac{1}{2}\times OC\times I$ we get,

$\frac{\sqrt{35}}{2}=\frac{1}{2}\times \sqrt{5}\times I\Rightarrow I=\sqrt{7}$

And $OP=OQ=\sqrt{O{C}^2+I^2}=\sqrt{12}$

So, $a_1^2+b_1^2+a_2^2+b_2^2=O{P}^2+O{Q}^2=12+12=24$