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 $z_{1} \cdot z_{2}=$
Let $z_{1}=2(\cos \alpha+\mathrm{i} \sin \alpha), z_{2}=3(\cos \beta+\mathrm{i} \sin \beta)$, then we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ 6 ( \cos \alpha - \cos \beta ) = \frac { 3 } { 2 } , } \\
{ 6 ( \sin \alpha - \sin \beta ) = - 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\cos \alpha-\cos \beta=-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}=\frac{1}{4}, \\
\sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}=-\frac{1}{6}
\end{array}\right.\right. \\
\Rightarrow \tan \frac{\alpha+\beta}{2}=\frac{3}{2} \Rightarrow\left\{\begin{array}{l}
\cos (\alpha+\beta)=\frac{1-\frac{9}{4}}{1+\frac{9}{4}}=-\frac{5}{13}, \\
\sin (\alpha+\beta)=\frac{3}{1+\frac{9}{4}}=\frac{12}{13} .
\end{array}\right.
\end{array}
$$

Therefore, $z_{1} \cdot z_{2}=6[\cos (\alpha+\beta)+\mathrm{i} \sin (\alpha+\beta)]=-\frac{30}{13}+\frac{72}{13} \mathrm{i}$.