97 If $\cos A+\cos B+\cos C=0$, then the value of $\frac{\cos 3 A+\cos 3 B+\cos 3 C}{\cos A \cos B \cos C}$ is $\qquad$
97 12. Since
$$
\cos A+\cos B+\cos C=0,
$$

then
$$
\cos ^{3} A+\cos ^{3} B+\cos ^{3} C=3 \cos A \cos B \cos C .
$$

Therefore, we have
$$
\begin{aligned}
& \frac{\cos 3 A+\cos 3 B+\cos 3 C}{\cos A \cos B \cos C} \\
= & \frac{4\left(\cos ^{3} A+\cos ^{3} B+\cos ^{3} C\right)-3(\cos A+\cos B+\cos C)}{\cos A \cos B \cos C} \\
= & \frac{4 \times 3 \cos A \cos B \cos C}{\cos A \cos B \cos C} \\
= & 12 .
\end{aligned}
$$