11. (This question is worth 20 points) In the plane rectangular coordinate system $x O y$, let $A B$ be a chord of the parabola $y^{2}=4 x$ passing through the point $F(1,0)$, and the circumcircle of $\triangle A O B$ intersects the parabola at point $P$ (different from points $O, A, B$). If $P F$ bisects $\angle A P B$, find all possible values of $|P F|$.
Let $A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right), P\left(\frac{y_{3}^{2}}{4}, y_{3}\right)$, and by the given conditions, $y_{1}, y_{2}, y_{3}$ are pairwise distinct and non-zero.
Assume the equation of line $A B$ is $x=t y+1$, and combining it with the parabola equation, we get $y^{2}-4 t y-4=0$, hence
$$
y_{1} y_{2}=-4 \text {. }
$$

Noting that the circumcircle of $\triangle A O B$ passes through point $O$, we can set the equation of the circle as $x^{2}+y^{2}+d x+e y=0$, and combining it with $x=\frac{y^{2}}{4}$, we get $\frac{y^{4}}{16}+\left(1+\frac{d}{4}\right) y^{2}+e y=0$. This quartic equation has four distinct real roots $y=y_{1}, y_{2}, y_{3}, 0$, so by Vieta's formulas, we have $y_{1}+y_{2}+y_{3}+0=0$, thus
$$
y_{3}=-\left(y_{1}+y_{2}\right) \text {. }
$$
$\qquad$
Since $P F$ bisects $\angle A P B$, by the Angle Bisector Theorem, we have $\frac{|P A|}{|P B|}=\frac{|F A|}{|F B|}=\frac{\left|y_{1}\right|}{\left|y_{2}\right|}$, combining (1) and (2), we get
$$
\begin{aligned}
\frac{y_{1}^{2}}{y_{2}^{2}} & =\frac{|P A|^{2}}{|P B|^{2}}=\frac{\left(\frac{y_{3}^{2}}{4}-\frac{y_{1}^{2}}{4}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}}{\left(\frac{y_{3}^{2}}{4}-\frac{y_{2}^{2}}{4}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}=\frac{\left(\left(y_{1}+y_{2}\right)^{2}-y_{1}^{2}\right)^{2}+16\left(2 y_{1}+y_{2}\right)^{2}}{\left(\left(y_{1}+y_{2}\right)^{2}-y_{2}^{2}\right)^{2}+16\left(2 y_{2}+y_{1}\right)^{2}} \\
& =\frac{\left(y_{2}^{2}-8\right)^{2}+16\left(4 y_{1}^{2}+y_{2}^{2}-16\right)}{\left(y_{1}^{2}-8\right)^{2}+16\left(4 y_{2}^{2}+y_{1}^{2}-16\right)}=\frac{y_{2}^{4}+64 y_{1}^{2}-192}{y_{1}^{4}+64 y_{2}^{2}-192}, \quad \ldots \ldots \cdots \cdots \cdots \cdots \cdots \cdots \cdots
\end{aligned}
$$

Thus, $y_{1}^{6}+64 y_{1}^{2} y_{2}^{2}-192 y_{1}^{2}=y_{2}^{6}+64 y_{2}^{2} y_{1}^{2}-192 y_{2}^{2}$, hence
$$
\left(y_{1}^{2}-y_{2}^{2}\right)\left(y_{1}^{4}+y_{1}^{2} y_{2}^{2}+y_{2}^{4}-192\right)=0 \text {. }
$$

When $y_{1}^{2}=y_{2}^{2}$, $y_{2}=-y_{1}$, so $y_{3}=0$, which means $P$ coincides with $O$, contradicting the given conditions.
When $y_{1}^{4}+y_{1}^{2} y_{2}^{2}+y_{2}^{4}-192=0$, noting (1), we have
$$
\left(y_{1}^{2}+y_{2}^{2}\right)^{2}=192+\left(y_{1} y_{2}\right)^{2}=208 \text {. }
$$

Since $y_{1}^{2}+y_{2}^{2}=4 \sqrt{13}>8=\left|2 y_{1} y_{2}\right|$, there exist real numbers $y_{1}, y_{2}$ satisfying (1) and $y_{1}^{2}+y_{2}^{2}=4 \sqrt{13}$, corresponding to points $A, B$ that meet the conditions. At this time, combining (1) and (2), we have
$$
|P F|=\frac{y_{3}^{2}}{4}+1=\frac{\left(y_{1}+y_{2}\right)^{2}+4}{4}=\frac{y_{1}^{2}+y_{2}^{2}-4}{4}=\frac{\sqrt{208}-4}{4}=\sqrt{13}-1 \text {. }
$$