Let $O$ be the center of the circle with radius $5\sqrt{2}$. Consider the circle with diameter $\overline{AO}$. Because $\angle ABO$ and $\angle ACO$ are right angles, the opposite angles of quadrilateral $ABOC$ are supplementary, and hence this quadrilateral is cyclic. Thus $O$ is also on the circle that passes through $A$, $B$, and $C$, and by symmetry $\overline{AO}$ is a diameter. By the Pythagorean Theorem,
$$AO = \sqrt{(5\sqrt{2})^2 + (3\sqrt{6})^2} = 2\sqrt{26},$$
so the circle that passes through $A$, $B$, and $C$ has radius $\sqrt{26}$ and area $26\pi$.