7. In $\triangle A B C$, the three interior angles $A, B, C$ satisfy: $A=3 B=9 C$, then $\cos A \cos B+\cos B \cos C+\cos C \cos A=$
$$
\begin{array}{l}
A=\frac{9 \pi}{13}, B=\frac{3 \pi}{13}, C=\frac{\pi}{13}, \text { then } \cos A \cos B+\cos B \cos C+\cos C \cos A \\
=\cos \frac{9 \pi}{13} \cos \frac{3 \pi}{13}+\cos \frac{3 \pi}{13} \cos \frac{\pi}{13}+\cos \frac{\pi}{13} \cos \frac{9 \pi}{13} \\
=\frac{1}{2}\left(\cos \frac{12 \pi}{13}+\cos \frac{6 \pi}{13}+\cos \frac{4 \pi}{13}+\cos \frac{2 \pi}{13}+\cos \frac{10 \pi}{13}+\cos \frac{8 \pi}{13}\right) \\
=\frac{1}{4 \sin \frac{\pi}{13}}\left(\sin \pi-\sin \frac{\pi}{13}\right)=-\frac{1}{4} .
\end{array}
$$