9. In $\triangle A B C$, it is known that $\overrightarrow{A B} \cdot \overrightarrow{A C}+2 \overrightarrow{B A} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{C B}$. Find the maximum value of $\sin C$.
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{A C}+2 \overrightarrow{B A} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{C B} \Rightarrow \frac{b^{2}+c^{2}-a^{2}}{2}+a^{2}+c^{2}-b^{2}=\frac{3\left(a^{2}+b^{2}-c^{2}\right)}{2} \\
\Rightarrow b^{2}+c^{2}-a^{2}+2\left(a^{2}+c^{2}-b^{2}\right)=3\left(a^{2}+b^{2}-c^{2}\right) \Rightarrow a^{2}+2 b^{2}=3 c^{2} .
\end{array}
$$

Thus, $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{3\left(a^{2}+b^{2}\right)-\left(a^{2}+2 b^{2}\right)}{6 a b}=\frac{2 a^{2}+b^{2}}{6 a b} \geqslant \frac{\sqrt{2}}{3}$,
equality holds when $\left\{\begin{array}{l}2 a^{2}=b^{2}, \\ a^{2}+2 b^{2}=3 c^{2}\end{array} \Rightarrow\left\{\begin{array}{l}b=\sqrt{2} a, \\ c=\frac{\sqrt{15}}{3} a \text {. }\end{array}\right.\right.$
At this point, $\sin C=\sqrt{1-\left(\frac{\sqrt{2}}{3}\right)^{2}}=\frac{\sqrt{7}}{3}$, which means the maximum value of $\sin C$ is $\frac{\sqrt{7}}{3}$.