Three, the base of the triangular pyramid $S-ABC$ is a regular $\triangle ABC$, with the side length of this triangle being 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.
Three, let $C K \perp A B, L$ be the intersection of its extension with the sphere, then $S K$ is perpendicular to $A B$, and the center of the sphere lies in the plane of $\triangle L S C$. The problem is reduced to finding the radius $R$ of its circumcircle. Since $C K=2 \sqrt{3}, L C=\frac{8}{\sqrt{3}}$ (the diameter of the circumcircle of the equilateral $\triangle A B C$), $S K=\sqrt{15}$, according to the cosine rule, from $\triangle S K C$ and $\triangle I S C$ we find $\cos \angle S C I=\frac{1}{2 \sqrt{3}}, I S=\sqrt{\frac{67}{3}}$. Now using the formula $2 R=\frac{L S}{\sin \angle S C L}$, we find $R=\sqrt{\frac{67}{11}}$. Therefore, $S_{\text {real }}=4 \pi R^{2}=4 \pi \cdot \frac{67}{11}=\frac{268}{11} \pi$.