9. Given the coordinates of the three vertices of $\triangle A B C$ are $A(0,0), B(7,0), C(3,4)$, a line passing through the point $(6-2 \sqrt{2}, 3-\sqrt{2})$ intersects line segments $A C, B C$ at points $P, Q$. If $S_{\triangle P Q C}=\frac{14}{3}$, then $|C P|+|C Q|=$ $\qquad$ .
Given, $I$ is the incenter of $\triangle ABC$, then $\frac{AD}{DB}=\frac{b}{a}$
$\Rightarrow AD=\frac{bc}{a+b} \Rightarrow \frac{CI}{ID}=\frac{b}{\frac{bc}{a+b}}=\frac{a+b}{c}$
$\Rightarrow CI=\frac{a+b}{a+b+c} CD$. Also, $\overrightarrow{AD}=\frac{b}{a+b} \overrightarrow{AB}$
$\Rightarrow \overrightarrow{CD}=\frac{a}{a+b} \overrightarrow{CA}+\frac{b}{a+b} \overrightarrow{CB}$
$\Rightarrow \overrightarrow{CI}=\frac{a}{a+b+c} \overrightarrow{CA}+\frac{b}{a+b+c} \overrightarrow{CB}$. Let $\overrightarrow{CA}=\lambda \overrightarrow{CP}, \overrightarrow{CB}=\mu \overrightarrow{CQ}$,
thus $\overrightarrow{CI}=\frac{a \lambda}{a+b+c} \overrightarrow{CP}+\frac{b \mu}{a+b+c} \overrightarrow{CQ} \Rightarrow a \lambda+b \mu=a+b+c$.
Therefore, $|CP|+|CQ|=\frac{b}{\lambda}+\frac{a}{\mu}=\frac{a \lambda+b \mu}{\lambda \mu}=\frac{a+b+c}{\lambda \mu}=\frac{12+4 \sqrt{2}}{3}$.
Note: The data in this problem is incorrect, and there are no specific values for $\lambda, \mu$.