Let $$P(a_1, b_1)$$ and $$Q(a_2, b_2)$$ 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 ______.

We have a circle with center $$C(\sqrt{2}, \sqrt{3})$$, origin $$O$$, and points $$P(a_1, b_1)$$ and $$Q(a_2, b_2)$$ on the circle such that $$\vec{OC}$$ is perpendicular to both $$\vec{CP}$$ and $$\vec{CQ}$$.

We compute $$|OC| = \sqrt{(\sqrt{2})^2 + (\sqrt{3})^2} = \sqrt{2 + 3} = \sqrt{5}$$.

Since $$\vec{OC}\perp \vec{CP}$$ and $$\vec{OC}\perp \vec{CQ}$$, both $$P$$ and $$Q$$ lie on the line through $$C$$ perpendicular to $$\vec{OC}$$.

Because $$P$$ and $$Q$$ lie on the circle centered at $$C$$, we have $$|CP| = |CQ| = r$$, and $$P$$ and $$Q$$ are symmetric about $$C$$ on this perpendicular line.

The area of triangle $$OCP$$ is $$\frac{1}{2}\cdot |OC|\cdot |CP| = \frac{1}{2}\cdot \sqrt{5}\cdot r = \frac{\sqrt{35}}{2}$$.

Equating gives $$\sqrt{5}\cdot r = \sqrt{35}$$, so $$r = \sqrt{7}$$.

By the Pythagorean theorem, since $$\vec{OC}\perp \vec{CP}$$, we have $$|OP|^2 = |OC|^2 + |CP|^2 = 5 + 7 = 12$$ and similarly $$|OQ|^2 = |OC|^2 + |CQ|^2 = 5 + 7 = 12$$.

Hence $$a_1^2 + b_1^2 = |OP|^2 = 12$$ and $$a_2^2 + b_2^2 = |OQ|^2 = 12$$, which yields $$a_1^2 + a_2^2 + b_1^2 + b_2^2 = 12 + 12 = 24$$.

The answer is $$\boxed{24}$$.