jmc

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 $x=\frac{6\pm \sqrt{(-6{)}^2-4(-3)(1)}}{2(1)}=\frac{6\pm \sqrt{48}}{2}=3\pm 2\sqrt{3}$. 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 $3+2\sqrt{3}\approx 6.46\dots$ is greater than 6, as we said it must be.)