9. (This question is worth 16 points) 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 $\sin C$.
Solution: By the definition of the dot product and the cosine rule, we have $\overrightarrow{A B} \cdot \overrightarrow{A C}=c b \cos A=\frac{b^{2}+c^{2}-a^{2}}{2}$.
Similarly, we get $\overrightarrow{B A} \cdot \overrightarrow{B C}=\frac{a^{2}+c^{2}-b^{2}}{2}, \overrightarrow{C A} \cdot \overrightarrow{C B}=\frac{a^{2}+b^{2}-c^{2}}{2}$. Therefore, the given condition can be transformed into
$$
b^{2}+c^{2}-a^{2}+2\left(a^{2}+c^{2}-b^{2}\right)=3\left(a^{2}+b^{2}-c^{2}\right),
$$

which simplifies to $a^{2}+2 b^{2}=3 c^{2}$.
By the cosine rule and the basic inequality, we get
$$
\begin{aligned}
\cos C & =\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{a^{2}+b^{2}-\frac{1}{3}\left(a^{2}+2 b^{2}\right)}{2 a b} \\
& =\frac{a}{3 b}+\frac{b}{6 a} \geq 2 \sqrt{\frac{a}{3 b} \cdot \frac{b}{6 a}}=\frac{\sqrt{2}}{3},
\end{aligned}
$$

Therefore,
$$
\sin C=\sqrt{1-\cos ^{2} C} \leq \frac{\sqrt{7}}{3},
$$

with equality holding if and only if $a: b: c=\sqrt{3}: \sqrt{6}: \sqrt{5}$. Thus, the maximum value of $\sin C$ is $\frac{\sqrt{7}}{3}$. 12 points $\qquad$