smc

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