In $\triangle A B C$,
$$
\frac{\overrightarrow{A B} \cdot \overrightarrow{B C}}{3}=\frac{\overrightarrow{B C} \cdot \overrightarrow{C A}}{2}=\frac{\overrightarrow{C A} \cdot \overrightarrow{A B}}{1} \text {. }
$$

Then $\tan A=$ $\qquad$ .
- 1. $\sqrt{11}$.

From the given information,
$$
\begin{array}{l}
\frac{a^{2}+c^{2}-b^{2}}{3}=\frac{b^{2}+a^{2}-c^{2}}{2}=\frac{c^{2}+b^{2}-a^{2}}{1} \\
=\frac{a^{2}+b^{2}+c^{2}}{6}=\frac{2 b^{2}}{3}=\frac{2 c^{2}}{4}=\frac{2 a^{2}}{5} \\
\Rightarrow a^{2}: b^{2}: c^{2}=5: 3: 4 \\
\Rightarrow \cos A=\frac{\sqrt{3}}{6} \\
\Rightarrow \tan A=\sqrt{11} .
\end{array}
$$