3. If the complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=2,\left|z_{2}\right|=3$, $3 z_{1}-2 z_{2}=\frac{3}{2}-\mathrm{i}$, then $z_{1} z_{2}=$ $\qquad$ .
$3.6\left(-\frac{5}{13}+\frac{12}{13} i\right)$.
Let $3 z_{1}=6(\cos \alpha+i \sin \alpha)$,
$2 z_{2}=6(\cos \beta+i \sin \beta)$.
Then $6(\cos \alpha-\cos \beta)=\frac{3}{2}, 6(\sin \alpha-\sin \beta)=-1$.
Thus $\tan \frac{\alpha+\beta}{2}=\frac{3}{2}, \cos (\alpha+\beta)=-\frac{5}{13}, \sin (\alpha+\beta)=\frac{12}{13}$.
Therefore, $z_{1} z_{2}=6\left(-\frac{5}{13}+\frac{12}{13} i\right)$.