As shown in Figure 3, from a point $P$ outside circle $\odot O$, two secants $PAB$ and $PCD$ are drawn, intersecting $\odot O$ at points $A, B$ and $C, D$ respectively. Chords $AD$ and $BC$ intersect at point $Q$, and $PQ$ intersects $\odot O$ at point $E$. If $PE = \frac{12}{5}$, the radius of $\odot O$ is 3, and the distance between points $O$ and $P$ is 5, find the length of $QE$.
Solution: As shown in Figure 3, draw $PT$ tangent to $\odot O$ at point $T$, and connect $OP, OT$. Then
$$
OP=5, OT=3, PT \perp OT \text{. }
$$

Thus, $PT=\sqrt{OP^2-OT^2}=\sqrt{5^2-3^2}=4$.
Extend $PQ$ to intersect $\odot O$ at point $F$.
Then $PT^2=PE \cdot PF \Rightarrow 4^2=\frac{12}{5} PF \Rightarrow PF=\frac{20}{3}$.
Since $\angle AQP > \angle PDQ$,
$\angle PDQ = \angle PBQ$,
therefore, $\angle AQP > \angle PBQ$.
Construct $\angle PBM = \angle AQP$, with $BM$ intersecting $QF$ at point $M$, then $A, B, M, Q$ are concyclic.
Thus, $\angle PAQ = \angle PMB$,
$PA \cdot PB = PQ \cdot PM = PE \cdot PF$.
It is easy to see that $\angle PAQ = \angle PCB$, therefore,
$\angle PMB = \angle PCB$.
Thus, $P, B, M, C$ are concyclic, and
$$
PQ \cdot QM = BQ \cdot QC = QE \cdot QF \text{. }
$$

Therefore, $PE \cdot PF - QE \cdot QF$
$$
\begin{array}{l}
= PQ \cdot PM - PQ \cdot QM \\
= PQ(PM - QM) = PQ^2 \\
\Rightarrow PE \cdot PF - (PQ - PE)(PF - PQ) = PQ^2 \\
\Rightarrow PQ \cdot PF + PQ \cdot PE = 2PE \cdot PF \\
\Rightarrow PQ = \frac{2PE \cdot PF}{PE + PF} = \frac{60}{17} \text{. }
\end{array}
$$

Thus, $QE = PQ - PE = \frac{60}{17} - \frac{12}{5} = \frac{96}{85}$.