3. Let the complex numbers $z_{1}=(2-a)+(1-b) \mathrm{i}, z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, z_{3}=(3-a)+(3-2 b) \mathrm{i}$, where $a, b \in \mathbf{R}$. When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ is minimized, $3 a+4 b=$ $\qquad$ .
3. 12 .

It is easy to find that $z_{1}+z_{2}+z_{3}=8+6 \mathrm{i}$, thus $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right|=10$. $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ achieves its minimum value if and only if
$$
\frac{2-a}{1-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{8}{6}
$$

Solving this, we get $a=\frac{7}{3}, b=\frac{5}{4}$. Therefore, $3 a+4 b=12$.