7. Given that the circumcenter of $\triangle A B C$ is $O$, and
$$
\overrightarrow{A O} \cdot \overrightarrow{B C}+2 \overrightarrow{B O} \cdot \overrightarrow{C A}+3 \overrightarrow{C O} \cdot \overrightarrow{A B}=0
$$

then the minimum value of $\cot A+\cot C$ is $\qquad$
7. $\frac{2 \sqrt{3}}{3}$.

Let $B C=a, C A=b, A B=c$. Draw $O D \perp B C$, with the foot of the perpendicular being $D$. It is easy to see that $D$ is the midpoint of side $B C$, so we have
$$
\overrightarrow{A D}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \text {. }
$$

Then $\overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C}=\overrightarrow{A D} \cdot \overrightarrow{B C}$
$$
\begin{array}{l}
=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \\
=\frac{1}{2}\left(|\overrightarrow{A C}|^{2}-|\overrightarrow{A B}|^{2}\right) \\
=\frac{1}{2}\left(b^{2}-c^{2}\right) .
\end{array}
$$

Thus, $\overrightarrow{A O} \cdot \overrightarrow{B C}+2 \overrightarrow{B O} \cdot \overrightarrow{C A}+3 \overrightarrow{C O} \cdot \overrightarrow{A B}=0$
$$
\Leftrightarrow \frac{1}{2}\left(b^{2}-c^{2}\right)+2 \times \frac{1}{2}\left(c^{2}-a^{2}\right)+3 \times \frac{1}{2}\left(a^{2}-b^{2}\right)=0 \text {. }
$$

Simplifying, we get $a^{2}+c^{2}=2 b^{2}$.
From $2 b^{2}=a^{2}+c^{2} \geqslant 2 a c$, we have $b^{2} \geqslant a c$.
Then $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{b^{2}}{2 a c} \geqslant \frac{b^{2}}{2 b^{2}}=\frac{1}{2}$
$$
\Rightarrow 0<\angle B \leqslant \frac{\pi}{3} \Rightarrow 0<\sin B \leqslant \frac{\sqrt{3}}{2} \text {. }
$$

Therefore, $\cot A+\cot C$
$$
\begin{array}{l}
=\frac{\sin A \cdot \cos C+\sin C \cdot \cos A}{\sin A \cdot \sin C} \\
=\frac{\sin B}{\sin A \cdot \sin C} \\
\geqslant \frac{1}{\sin B} \geqslant \frac{2 \sqrt{3}}{3} .
\end{array}
$$

The equality holds if and only if $\angle A=\angle B=\angle C=\frac{\pi}{3}$.
Thus, the minimum value of $\cot A+\cot C$ is $\frac{2 \sqrt{3}}{3}$.