97 Given $z_{1}=x+\sqrt{5}-y i, z_{2}=x-\sqrt{5}+y i$. Here $x, y$ are real numbers, and $\left|z_{1}\right|+\left|z_{2}\right|=6$, then
$$
f(x, y)=|2 x-3 y-12|
$$

the minimum value is ( )
A. 0
B. $12+6 \sqrt{2}$
C. $12-6 \sqrt{2}$
D. $2-\sqrt{2}$
97 C. From $\left|z_{1}\right|+\left|z_{2}\right|=6$ we get
$$
\sqrt{(x+\sqrt{5})^{2}+y^{2}}+\sqrt{(x-\sqrt{5})^{2}+y^{2}}=6,
$$

Thus, the point $(x, y)$ is the locus of points whose sum of distances to the two fixed points $(\sqrt{5}, 0)$ and $(-\sqrt{5}, 0)$ is 6, hence $(x, y)$ satisfies the ellipse equation
$$
\frac{x^{2}}{9}+\frac{y^{2}}{4}=1
$$

Let $x=3 \cos \theta, y=2 \sin \theta$, substituting into $f(x, y)=|2 x-3 y-12|$, we get
$$
f(\theta)=|6 \cos \theta-6 \sin \theta-12|=6\left|\sqrt{2} \cos \left(\theta+\frac{\pi}{4}\right)-2\right|
$$

When $\theta=\frac{\pi}{4}$, $f(x, y)$ attains its minimum value $6|\sqrt{2}-2|=12-6 \sqrt{2}$.