11. As shown in the figure, 12 congruent disks are placed on the circumference of a circle $C$ with radius 1, such that they cover $C$. No two disks overlap, and adjacent disks are tangent to each other. The sum of the areas of these disks can be written as
$$
\pi(a-b \sqrt{c}) \text {, where }
$$
$a, b, c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$.
11. By the people's knowledge, 6-certainly passes through the tangency points of each circle, and each tangent line passes through the center $O$ of circle $C$ (as shown in the figure).
$P$ is the center of a circle,
$$
\begin{array}{l}
\angle P B O=90^{\circ}, \\
\angle B O A=\frac{1}{12} 2 \pi=\frac{\pi}{6}, \\
\angle P O B=\frac{\pi}{12} .
\end{array}
$$

Therefore, $P B=B O$
- $\operatorname{tg} \angle P O B=\operatorname{tg} \frac{\pi}{12}$
$$
\begin{array}{l}
=\operatorname{tg}\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\frac{\ln \frac{\pi}{3} \cdots \operatorname{tg} \frac{\pi}{4}}{1+\operatorname{tg} \frac{\pi}{3} \operatorname{tg} \frac{\pi}{4}} \\
=\frac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3} .
\end{array}
$$

Thus, the area of each disk is $\pi(2-\sqrt{3})^{2}=$ $\pi(7-4 \sqrt{ } \overline{3})$. The sum of the areas of these disks is
$$
\begin{array}{l}
12 \pi(7-4 \sqrt{3})=\pi(84-48 \sqrt{3}) . \\
\text { Hence } a+b+c=84+48+j=135 .
\end{array}
$$