Example 2 In an isosceles right $\triangle ABC$, it is known that $\angle ACB=$ $90^{\circ}, P$ is a point inside $\triangle ABC$, such that $PA=11, PB=7, PC=$ 6. Then the length of side $AC$ is $\qquad$ [2]
As shown in Figure 3, $\triangle P B C$ is rotated $90^{\circ}$ clockwise around point $C$, making point $B$ coincide with $A$, and point $P$ moves to position $P^{\prime}$, connecting $P P^{\prime}$.

Then $\triangle C P P^{\prime}$ is an isosceles right triangle. Thus,
$$
P P^{\prime}=\sqrt{P C^{2}+P^{\prime} C^{2}}=6 \sqrt{2} \text {. }
$$

By the properties of figure rotation, we have
$$
P^{\prime} A=P B=7 \text {. }
$$

Therefore, $P P^{\prime 2}+P^{\prime} A^{2}=(6 \sqrt{2})^{2}+7^{2}=121$.
And $P A^{2}=11^{2}=121$, so
$$
P P^{\prime 2}+P^{\prime} A^{2}=P A^{2} \text {. }
$$

Hence, $\triangle A P P^{\prime}$ is a right triangle, and $\angle A P^{\prime} P=90^{\circ}$.
Therefore, $\angle C P^{\prime} A=90^{\circ}+45^{\circ}=135^{\circ}$.
Draw $A Q \perp C P^{\prime}$, intersecting the extension of $C P^{\prime}$ at point
$Q$. Then
$$
\begin{array}{l}
\angle Q P^{\prime} A=\angle Q A P^{\prime}=45^{\circ} \Rightarrow 2 A Q^{2}=P^{\prime} A^{2} \\
\Rightarrow A Q=P^{\prime} Q=\frac{7 \sqrt{2}}{2} \Rightarrow C Q=6+\frac{7 \sqrt{2}}{2} .
\end{array}
$$

Thus, $A C^{2}=A Q^{2}+C Q^{2}=85+42 \sqrt{2}$.
Therefore, $A C=\sqrt{85+42 \sqrt{2}}$.