64 The base of the triangular pyramid $S-ABC$ is an equilateral triangle $ABC$, with a side length of 4. It is also known that $AS=BS=\sqrt{19}$, and $CS=3$. Find the surface area of the circumscribed sphere of this triangular pyramid.
64 As shown in Figure 29 (a), let the midpoint of $AB$ be $E$, then the center of the circumscribed sphere $O$ is in $\triangle SCE$. In $\triangle SCE$,
$$
\begin{array}{c}
SE=\sqrt{SA^{2}-\left(\frac{AB}{2}\right)^{2}}=\sqrt{15}, SC=3 . \\
CE=\frac{\sqrt{3}}{2} \times AB=2 \sqrt{3} . \text{ As shown in Figure } 29 \text{ (b) }
\end{array}
$$

The center $O$ is on the perpendicular bisector of $SC$, and also on the perpendicular to $EC$, which passes through the point $G$ (the centroid of $\triangle ABC$). Extend $CE$ to $F$ such that $EF=\frac{1}{3} CE=\frac{2}{\sqrt{3}}$, then $O$ is the circumcenter of $\triangle SCF$.
$$
\begin{array}{l}
\cos C=\frac{3^{2}+(2 \sqrt{3})^{2}-(\sqrt{15})^{2}}{2 \times 3 \times 2 \sqrt{3}}=\frac{1}{2 \sqrt{3}} \\
\sin C=\sqrt{1-\left(\frac{1}{2 \sqrt{3}}\right)^{2}}=\sqrt{\frac{11}{12}} . \\
SF^{2}=CS^{2}+CF^{2}-2 \times CS \times CF \times \cos C \\
=9+\frac{64}{3}-2 \times 3 \times \frac{8}{\sqrt{3}} \times \frac{1}{2 \sqrt{3}} \\
=\frac{61}{3} \text {. } \\
\end{array}
$$

The circumradius $R=\frac{SF}{2 \sin C}=\sqrt{\frac{61}{3}} \cdot \frac{1}{2} \sqrt{\frac{12}{11}}=\sqrt{\frac{61}{11}}$
The surface area of the circumscribed sphere $=4 \pi R=4 \pi \sqrt{\frac{61}{11}}$.
Figure 29