9. Let the complex number $z_{1}=(6-a)+(4-b) \mathrm{i}$,
$$
\begin{array}{l}
z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, \\
z_{3}=(3-a)+(3-2 b) \mathrm{i},
\end{array}
$$

where, $a, b \in \mathbf{R}$. When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ achieves its minimum value, $3 a+4 b=$ $\qquad$
9.12.

It is easy to find that $z_{1}+z_{2}+z_{3}=12+9 \mathrm{i}$. Therefore,
$$
\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right|=15 \text {. }
$$

The equality holds if and only if $\frac{6-a}{4-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{12}{9}$. Solving this, we get $a=\frac{7}{3}, b=\frac{5}{4}$.
Thus, $3 a+4 b=12$.