3. Let $O$ be the circumcenter of $\triangle A B C$, and
$$
\overrightarrow{A O} \cdot \overrightarrow{B C}+2 \overrightarrow{B O} \cdot \overrightarrow{C A}+3 \overrightarrow{C O} \cdot \overrightarrow{A B}=0 \text {. }
$$

Then the minimum value of $\frac{1}{\tan A}+\frac{1}{\tan C}$ is
3. $\frac{2 \sqrt{3}}{3}$.

Let $B C=a, C A=b, A B=c, D$ be the midpoint of $B C$. Then $O D \perp B C$.
$$
\begin{array}{l}
\text { Hence } \overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C}=\overrightarrow{A D} \cdot \overrightarrow{B C} \\
=\frac{1}{2}(\overrightarrow{A C}+\overrightarrow{A B}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \\
=\frac{1}{2}\left(b^{2}-c^{2}\right) .
\end{array}
$$

Similarly, $\overrightarrow{B O} \cdot \overrightarrow{A C}=\frac{1}{2}\left(c^{2}-a^{2}\right)$,
$$
\overrightarrow{C O} \cdot \overrightarrow{A B}=\frac{1}{2}\left(a^{2}-b^{2}\right) \text {. }
$$

By the problem statement,
$$
\begin{array}{l}
\frac{1}{2}\left(b^{2}-c^{2}+2 c^{2}-2 a^{2}+3 a^{2}-3 b^{2}\right)=0 \\
\Rightarrow a^{2}+c^{2}=2 b^{2} \\
\Rightarrow \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{b^{2}}{2 a c} \\
\quad \geqslant \frac{b^{2}}{a^{2}+c^{2}}=\frac{1}{2} \\
\Rightarrow 0<\angle B \leqslant \frac{\pi}{3} \\
\Rightarrow b^{2} \geqslant a c .
\end{array}
$$

Therefore, $\frac{1}{\tan A}+\frac{1}{\tan C}$
$$
\begin{array}{l}
=\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}=\frac{\sin (A+C)}{\sin A \cdot \sin C} \\
=\frac{\sin B}{\sin A \cdot \sin C}=\frac{b^{2}}{a c \sin B} \\
\geqslant \frac{1}{\sin B} \geqslant \frac{1}{\sin \frac{\pi}{3}}=\frac{2 \sqrt{3}}{3},
\end{array}
$$

with equality holding if and only if $\triangle A B C$ is an equilateral triangle.