7. Given $P$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{P A}+\overrightarrow{P B}=\lambda \overrightarrow{P C}, \tan C$ $=\frac{12}{5}$, then the value of the real number $\lambda$ is $\qquad$.
Let the midpoint of $A B$ be $M$. Since $\overrightarrow{P A}+\overrightarrow{P B}=2 \overrightarrow{P M} \Rightarrow C, P, M$ are collinear, and $P$ is the circumcenter, then $C M \perp A B \Rightarrow C A=C B$. Given $\tan C=\frac{12}{5} \Rightarrow \tan \frac{C}{2}=\frac{2}{3} \Rightarrow M B=\frac{2}{3} C M$. Assume $C M=3, M B=2, P C=R \Rightarrow(3-R)^{2}+4=R^{2} \Rightarrow R=P C=\frac{13}{6}, P M=\frac{5}{6}$. Therefore, $2 \overrightarrow{P M}=\lambda \overrightarrow{P C} \Rightarrow \lambda=-\frac{2 P M}{P C}=-\frac{10}{13}$.