Question 147, Given that the base $ABC$ of the tetrahedron $S-ABC$ is an equilateral triangle, point $A$'s projection $H$ on the face $SBC$ is the orthocenter of $\triangle SBC$, the dihedral angle $H-AB-C$ is $30^{\circ}$, and $SA=2$, then the volume of this tetrahedron is $\qquad$ -
Answer: Draw $AH \perp SBC$ at $H$, draw $BH \perp SC$ at $E$, then $SC \perp AH, SC \perp BE$, so $SC \perp$ plane $ABE$, hence $SC \perp AB$. The projection of $S$ on plane $ABC$ is $O$, then $SO \perp$ plane $ABC$, by the theorem of three perpendiculars, we know $CO \perp AB$.

Similarly, $BO \perp AC$, so point $O$ is the orthocenter of $\triangle ABC$. Since $\triangle ABC$ is an equilateral triangle, $O$ is the center of $\triangle ABC$, thus $SA=SB=SC=2$.

Extend $CO$ to intersect $AB$ at $F$, then $F$ is the midpoint of $AB$, and $EF \perp AB$, so $\angle EFC$ is the plane angle of the dihedral angle $H-AB-C$, hence $\angle EFC=30^{\circ}$. Therefore, $\angle SCO=60^{\circ}$, so $SO=SC \cdot \sin 60^{\circ}=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}, CO=SC \cdot \cos 60^{\circ}=1$, thus $S_{\triangle ABC}=\frac{3 \sqrt{3}}{4}$. Hence $V_{S-ABC}=\frac{S_{\triangle ABC} \cdot SO}{3}=\frac{\frac{3 \sqrt{3}}{4} \times \sqrt{3}}{3}=\frac{3}{4}$.