6. 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 $\qquad$
$$
\text { 6. } 85+42 \sqrt{2} \text {. }
$$

Rotate $\triangle P B C$ 90 degrees clockwise around point $C$, so that point $B$ coincides with $A$, and point $P$ moves to position $P^{\prime}$ (as shown in Figure 4), and connect $P P^{\prime}$. Then $\triangle C P P^{\prime}$ is an isosceles right triangle, and $P P^{\prime}=6 \sqrt{2}$.
From $P P^{\prime 2}+P^{\prime} A^{2}$
$=72+49$
$$
=121=P A^{2} \text {, }
$$

we get $\angle P P^{\prime} A=90^{\circ}$.
Furthermore,
$\angle C P^{\prime} A=135^{\circ}$.
Draw $A Q \perp$
$C P^{\prime}$ at point $Q$. Then
$$
A Q=\frac{7}{\sqrt{2}}, C Q=6+\frac{7}{\sqrt{2}} \text {. }
$$

Thus, $A C^{2}=\left(6+\frac{7}{\sqrt{2}}\right)^{2}+\left(\frac{7}{\sqrt{2}}\right)^{2}=85+42 \sqrt{2}$.