Isosceles $\triangle ABC$ has a right angle at $C$.  Point $P$ is inside $\triangle ABC$, such that $PA=11$, $PB=7$, and $PC=6$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$, where $a$ and $b$ are positive integers.  What is $a+b$?

$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$
Using the Law of Cosines on $\triangle PBC$, we have:
\begin{align*} PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. \end{align*}
Using the Law of Cosines on $\triangle PAC$, we have:
\begin{align*} PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. \end{align*}
Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$.
\begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ &\Rightarrow s^4-170s^2+3697 = 0 \\ &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2  \end{align*}
Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$.  Thus, $a+b=85+42=\boxed{127}$.