Question 151, Given that all vertices of the tetrahedron $S-ABC$ are on the surface of sphere $O$, $\triangle \mathrm{ABC}$ is an equilateral triangle with side length 1, $SC$ is the diameter of sphere $O$, and $SC=2$, then the volume of the tetrahedron $S-ABC$ $V_{S-ABC}=$ $\qquad$ ـ.
Question 151, Solution: As shown in the figure, according to the conditions, the projection of 0 on plane $ABC$ is the circumcenter $P$ of $\triangle ABC$. Since $\mathrm{PC}=\frac{1}{\sqrt{3}}, \mathrm{OC}=1$, it follows that $\mathrm{OP}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$, and thus $\mathrm{d}(\mathrm{S},$ plane $\mathrm{ABC})=2 \mathrm{OP}=\frac{2 \sqrt{6}}{3}$. Therefore, $V_{S-ABC}=\frac{1}{3} \cdot S_{\triangle ABC} \cdot d(S$, plane $ABC)=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot \frac{2 \sqrt{6}}{3}=\frac{\sqrt{2}}{6}$