7. In the tetrahedron $SABC$, $\triangle ABC$ is an equilateral triangle. The projection of point $A$ onto the plane $SBC$ is $H$, which is the orthocenter of $\triangle SBC$. The dihedral angle $H-AB-C$ has a plane angle of $30^{\circ}$, and $SA=2\sqrt{3}$. What is the volume of the tetrahedron $SABC$?
7. $\frac{9}{4} \sqrt{3}$ As shown in the figure, given that $A H \perp$ plane $S B C$, connect $B H$ intersecting $S C$ at $E$, then $B H \perp S C$. By the theorem of three perpendiculars, $S C \perp A E$, hence $S C \perp$ plane $A B E$.
Let the projection of $S$ on plane $A B C$ be $O$, then $S O \perp$ plane $A B C$.
Since $S C \perp A B$, by the inverse of the theorem of three perpendiculars, $O O \perp A B$ at $F$.
Similarly, $B O \perp A C$, so $O$ is the orthocenter of $\triangle A B C$.
Since $\triangle A B C$ is an equilateral triangle, $O$ is the center of $\triangle A B C$,
thus $O A=O B=O C$,
so $S A=S B=S C=2 \sqrt{3}$.
Since $C F \perp A B$, and $C F$ is the projection of $E F$ on plane $A B C$, by the theorem of three perpendiculars, $E F \perp A B$, so $\angle E F C$ is the plane angle of the dihedral angle $H-A B-C$, hence $\angle E F C=30^{\circ}$,
$$
\begin{array}{l}
O C=S C \cdot \cos 60^{\circ}=2 \sqrt{3} \cdot \frac{1}{2}=\sqrt{3}, \\
S O=\sqrt{3} \cdot \tan 60^{\circ}=\sqrt{3} \cdot \sqrt{3}=3 .
\end{array}
$$

Also, $A B=\sqrt{3} \cdot O C=\sqrt{3} \cdot \sqrt{3}=3$.
Therefore, $V_{S-A B C}=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot 3^{2} \cdot 3=\frac{9}{4} \sqrt{3}$.