5 In $\triangle A B C$, if $\frac{\overrightarrow{A B} \cdot \overrightarrow{B C}}{3}=\frac{\overrightarrow{B C} \cdot \overrightarrow{C A}}{2}=\frac{\overrightarrow{C A} \cdot \overrightarrow{A B}}{1}$, then $\tan A=$
(5) $\sqrt{11}$ Hint: From the given,
$$
\frac{c^{2}+a^{2}-b^{2}}{3}=\frac{a^{2}+b^{2}-c^{2}}{2}=\frac{b^{2}+c^{2} \div a^{2}}{1},
$$

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