12. Given that the base of the tetrahedron $S-ABC$ is an equilateral triangle, the projection of point $A$ on the face $SBC$ is $H$, which is the orthocenter of $\triangle SBC$, the dihedral angle $H-AB-C$ is $30^{\circ}$, and $SA=2\sqrt{3}$. Then, the volume of the tetrahedron $S-ABC$ is $\qquad$.
Let the midpoint of $BC$ be $D$, and connect $AD, SD$. Then
$$
\left\{\begin{array}{l}
A H \perp B C, \\
A D \perp B C
\end{array} \Rightarrow B C \perp \text { plane } A H D \Rightarrow B C \perp H D\right.
$$
$\Rightarrow A, H, D$ are collinear and plane $ABC \perp$ plane $SAD$.
Connect $BH$ and extend it to intersect $SC$ at point $E$, then $\left\{\begin{array}{l}B H \perp S C, \\ A H \perp S C\end{array}\right.$
$\Rightarrow SC \perp$ plane $ABE \Rightarrow SC \perp AB$.
Draw $SO \perp AD \Rightarrow SO \perp$ plane $ABC \Rightarrow AB \perp$ plane $SCO \Rightarrow CO \perp AB$ $\Rightarrow O$ is the orthocenter of $\triangle ABC \Rightarrow O$ is the centroid of $\triangle ABC \Rightarrow S-ABC$ is a regular tetrahedron.
Extend $CO$ to intersect $AB$ at point $F \Rightarrow \angle EFC=30^{\circ} \Rightarrow \angle SCO=60^{\circ} \Rightarrow CO=\sqrt{3}$ $\Rightarrow AC=3, SO=3 \Rightarrow V_{S-ABC}=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} \cdot 9 \cdot 3=\frac{9 \sqrt{3}}{4}$.