10. The base of the tetrahedron $S-ABC$ is an equilateral triangle with side length 4. It is known that $AS=BS=\sqrt{19}, CS=3$. Find the surface area of the circumscribed sphere of the tetrahedron $S-ABC$.
10. Solution: Let the midpoint of $AB$ be $M$, and the center of the circumscribed sphere be $O$. Since $OA = OB$, we know that $OM$ is the perpendicular bisector of $AB$, and the plane $SMC$ is the perpendicular bisector plane of $AB$.
Since $OA = OB = OC$, we know that the projection of $O$ on $CM$, denoted as $O'$, is the centroid of $\triangle ABC$, satisfying $CO' = 2MO'$. It is easy to know that $SM = \sqrt{15}$, $SC = 3$, $CM = 2\sqrt{3}$. Let $S'$ be the projection of $S$ on $CM$. In $\triangle SMC$, $\cos \angle SMC = \frac{SM^2 + MC^2 - SC^2}{2 \cdot SM \cdot MC} = \frac{3}{2\sqrt{5}}$, so $\sin \angle SMC = \frac{\sqrt{11}}{2\sqrt{5}}$.
Thus, $SS' = SM \cdot \sin \angle SMC = \frac{\sqrt{33}}{2}$,
and $MS' = SM \cdot \cos \angle SMC = \frac{3\sqrt{3}}{2} \Rightarrow S'C = \frac{\sqrt{3}}{2}$.
Also, $MO' = \frac{2\sqrt{3}}{3}$, $O'C = \frac{4\sqrt{3}}{3}$, so $O'S' = O'C - S'C = \frac{5\sqrt{3}}{6}$.
Let $d = OO'$, and take the midpoint of $SC$ as $M$, then $OM \perp SC$.
By $S_{\triangle SOC} + S_{\triangle OO'C} = S_{\triangle SS'C} + S_{\text{quadrilateral } SOO'S'}$, $\Rightarrow \frac{1}{2} \cdot SC \cdot OM + \frac{1}{2} \cdot OO' \cdot O'C = \frac{1}{2} \cdot SS' \cdot S'C + \frac{1}{2} \left(SS' + OO'\right) \cdot O'S'$, $\Rightarrow 3 \sqrt{d^2 + \left(\frac{4\sqrt{3}}{3}\right)^2 - \left(\frac{3}{2}\right)^2} + \frac{4\sqrt{3}}{3} d = \frac{\sqrt{33}}{2} \cdot \frac{\sqrt{3}}{2} + \left(\frac{\sqrt{33}}{2} + d\right) \cdot \frac{5\sqrt{3}}{6}$, $\Rightarrow 2\sqrt{11} - \frac{\sqrt{3}}{2} d = 3 \sqrt{d^2 + \frac{37}{12}}$, solving for $d$ gives $d = \frac{5\sqrt{33}}{33}$. Therefore, $OC^2 = d^2 + O'C^2 = \frac{67}{11}$, so the surface area of the circumscribed sphere of the tetrahedron $S-ABC$ is $4\pi \cdot OC^2 = 4\pi \cdot \frac{67}{11} = \frac{268}{11} \pi$.