9. (16 points) In $\triangle A B C$, $\overrightarrow{A B} \cdot \overrightarrow{A C}+2 \overrightarrow{B A} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{C B}$.
Find the maximum value of $\sin C$.
From the definition of the dot product and the cosine rule, we have
$$
\overrightarrow{A B} \cdot \overrightarrow{A C}=c b \cos A=\frac{b^{2}+c^{2}-a^{2}}{2} \text {. }
$$

Similarly, $\overrightarrow{B A} \cdot \overrightarrow{B C}=\frac{a^{2}+c^{2}-b^{2}}{2}$,
$$
\overrightarrow{C A} \cdot \overrightarrow{C B}=\frac{a^{2}+b^{2}-c^{2}}{2} \text {. }
$$

Thus, the given equation can be rewritten as
$$
\begin{array}{l}
b^{2}+c^{2}-a^{2}+2\left(a^{2}+c^{2}-b^{2}\right)=3\left(a^{2}+b^{2}-c^{2}\right) \\
\Rightarrow a^{2}+2 b^{2}=3 c^{2} .
\end{array}
$$

Using the cosine rule and the basic inequality, we get
$$
\begin{array}{l}
\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{a^{2}+b^{2}-\frac{1}{3}\left(a^{2}+2 b^{2}\right)}{2 a b} \\
=\frac{a}{3 b}+\frac{b}{6 a} \geqslant 2 \sqrt{\frac{a}{3 b} \cdot \frac{b}{6 a}}=\frac{\sqrt{2}}{3} \\
\Rightarrow \sin C=\sqrt{1-\cos ^{2} C} \leqslant \frac{\sqrt{7}}{3},
\end{array}
$$

with equality holding if and only if $a: b: c=\sqrt{3}: \sqrt{6}: \sqrt{5}$. 
Therefore, the maximum value of $\sin C$ is $\frac{\sqrt{7}}{3}$.