Let $ x=6+\frac{1}{2+\frac{1}{6+\frac{1}{2+\frac{1}{6+\cdots}}}} $. Then we have $ x=6+\frac{1}{2+\frac{1}{x}} $. This means $ x-6=\frac{1}{2+\frac{1}{x}} $ or $ (x-6)\left(2+\frac{1}{x}\right)=1 $. Expanding the product gives $ 2x-12+1-\frac{6}{x}=1 $, or $ 2x-12-\frac{6}{x}=0 $. Multiply through by $ x $ and divide by $ 2 $ to find $ x^2-6x-3=0 $. Using the quadratic formula, we find $ x=\frac{6\pm\sqrt{(-6)^2-4(-3)(1)}}{2(1)}=\frac{6\pm\sqrt{48}}{2}=3\pm2\sqrt{3} $. Looking at the original expression for $ x $, we can see that it is greater than $ 6 $. So we take the positive value $ 3+2\sqrt{3} $ and get $ a+b+c=3+2+3=\boxed{8} $. (Remark: Notice that $ 3+2\sqrt{3}\approx 6.46\ldots $ is greater than 6, as we said it must be.)