2. Given a tetrahedron $P-ABC$ with the base $\triangle ABC$ being an equilateral triangle with side length 6, and $PA \perp$ plane $ABC$, $PA = 4$. If point $Q$ satisfies
$$
\overrightarrow{PQ}=\frac{1}{2}(\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}),
$$

then the volume of the tetrahedron $Q-ABC$ is $\qquad$
2. $6 \sqrt{3}$.

Let the center of $\triangle ABC$ be $O$.
Then $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=0$
$$
\begin{aligned}
\Rightarrow & \overrightarrow{P Q}=\frac{1}{2}(\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}) \\
& =\frac{1}{2}[(\overrightarrow{P O}+\overrightarrow{O A})+(\overrightarrow{P O}+\overrightarrow{O B})+(\overrightarrow{P O}+\overrightarrow{O C})] \\
& =\frac{3}{2} \overrightarrow{P O} \\
\Rightarrow & \overrightarrow{P O}=2 \overrightarrow{O Q} .
\end{aligned}
$$
$$
\begin{array}{l}
=\frac{1}{2} \times \frac{1}{3} S_{\triangle A H C} \cdot P A \\
=\frac{1}{2} \times \frac{1}{3} \times 9 \sqrt{3} \times 4=6 \sqrt{3} .
\end{array}
$$