60 Given $z_{1}=x+\sqrt{5}+y i, z_{2}=x-\sqrt{5}+y i$, and $x, y \in \mathbf{R},\left|z_{1}\right|+\left|z_{2}\right|=6$.
$f(x, y)=|2 x-3 y-12|$ The product of the maximum and minimum values of is $\qquad$
60 72. From $\left|z_{1}\right|+\left|z_{2}\right|=6$, we get $|x+y i+\sqrt{5}|+|x+y i-\sqrt{5}|=6$, which indicates that $(x, y)$ are points on the ellipse $4 x^{2}+9 y^{2}=36$. Let $x=3 \cos \theta, y=2 \sin \theta$, then
$$
\begin{aligned}
f(x, y) & =|2 x-3 y-12|=|6 \cos \theta-6 \sin \theta-12| \\
& =6\left|\sqrt{2} \cos \left(\theta+\frac{\pi}{4}\right)-2\right| .
\end{aligned}
$$

Therefore,
$$
f(x, y)_{\min }=12-6 \sqrt{2}, f(x, y)_{\max }=12+6 \sqrt{2} .
$$