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

then $\tan A=$ $\qquad$
4. $\sqrt{11}$.

From the given information,
$$
\begin{array}{l}
\frac{c^{2}+a^{2}-b^{2}}{3}=\frac{a^{2}+b^{2}-c^{2}}{2} \\
=\frac{b^{2}+c^{2}-a^{2}}{1}=\frac{a^{2}+b^{2}+c^{2}}{6} .
\end{array}
$$

Therefore, $a^{2}: b^{2}: c^{2}=5: 3: 4$.
By the cosine rule,
$$
\cos A=\frac{1}{2 \sqrt{3}} \Rightarrow \tan A=\sqrt{11} .
$$