7. Let the three interior angles $\angle A, \angle B, \angle C$ of $\triangle ABC$ satisfy $\angle A=3 \angle B=9 \angle C$. Then
$$
\begin{array}{l}
\cos A \cdot \cos B+\cos B \cdot \cos C+\cos C \cdot \cos A \\
=\quad .
\end{array}
$$
7. $-\frac{1}{4}$.

Let $\angle C=\theta, \angle B=3 \theta, \angle A=9 \theta$.
By $\theta+3 \theta+9 \theta=\pi \Rightarrow \theta=\frac{\pi}{13}$.
Let $S=\cos A \cdot \cos B+\cos B \cdot \cos C+\cos C \cdot \cos A$.
$$
\begin{array}{c}
\text { Then } S=\cos \frac{9 \pi}{13} \cdot \cos \frac{3 \pi}{13}+\cos \frac{3 \pi}{13} \cdot \cos \frac{\pi}{13}+ \\
\cos \frac{\pi}{13} \cdot \cos \frac{9 \pi}{13} \\
=\frac{1}{2}\left(\left(\cos \frac{12 \pi}{13}+\cos \frac{6 \pi}{13}\right)+\left(\cos \frac{4 \pi}{13}+\cos \frac{2 \pi}{13}\right)+\right. \\
\left.\left(\cos \frac{10 \pi}{13}+\cos \frac{8 \pi}{13}\right)\right) .
\end{array}
$$

Notice that, the angles in the parentheses form an arithmetic sequence with a common difference of $\frac{2 \pi}{13}$.
Multiplying both sides by $4 \sin \frac{\pi}{13}$ gives
$$
\begin{array}{l} 
4 S \sin \frac{\pi}{13} \\
= 2 \sin \frac{\pi}{13}\left(\cos \frac{2 \pi}{13}+\cos \frac{4 \pi}{13}+\cos \frac{6 \pi}{13}+\right. \\
\left.\cos \frac{8 \pi}{13}+\cos \frac{10 \pi}{13}+\cos \frac{12 \pi}{13}\right) \\
=\left(\sin \frac{3 \pi}{13}-\sin \frac{\pi}{13}\right)+\left(\sin \frac{5 \pi}{13}-\sin \frac{3 \pi}{13}\right)+ \\
\left(\sin \frac{7 \pi}{13}-\sin \frac{5 \pi}{13}\right)+\left(\sin \frac{9 \pi}{13}-\sin \frac{7 \pi}{13}\right)+ \\
\left(\sin \frac{11 \pi}{13}-\sin \frac{9 \pi}{13}\right)+\left(\sin \frac{13 \pi}{13}-\sin \frac{11 \pi}{13}\right) \\
=-\sin \frac{\pi}{13} \\
\Rightarrow S=-\frac{1}{4} .
\end{array}
$$