8. 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 $z_{1} \cdot z_{2}=$ $\qquad$
8. $-\frac{30}{13}+\frac{72}{13} \mathrm{i}$.

From
$$
\begin{array}{l}
3 z_{1}-2 z_{2}=\frac{1}{3} z_{2} \cdot \bar{z}_{2} \cdot z_{1}-\frac{1}{2} z_{1} \cdot \bar{z}_{1} \cdot z_{2} \\
=\frac{1}{6} z_{1} \cdot z_{2}\left(2 \bar{z}_{2}-3 \bar{z}_{1}\right)
\end{array}
$$

we can get $z_{1} \cdot z_{2}=\frac{6\left(3 z_{1}-2 z_{2}\right)}{2 \bar{z}_{2}-3 \bar{z}_{1}}=\frac{6\left(3 z_{1}-2 z_{2}\right)}{2 z_{2}-3 z_{1}}$
$$
=-6 \times \frac{\frac{3}{2}-\mathrm{i}}{\frac{3}{2}+\mathrm{i}}=-\frac{30}{13}+\frac{72}{13} \mathrm{i} .
$$

This problem can also be solved by setting up the complex numbers in trigonometric form.