5. If point $P$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{P A}+\overrightarrow{P B}+\lambda \overrightarrow{P C}=\mathbf{0}, C=120^{\circ}$, then the value of the real number $\lambda$ is $\qquad$.
Let $M$ be the midpoint of $AB$, then $\overrightarrow{P A}+\overrightarrow{P B}+\lambda \overrightarrow{P C}=2 \overrightarrow{P M}+\lambda \overrightarrow{P C}=\mathbf{0} \Rightarrow P, M, C$ are collinear, and $P M \perp A B \Rightarrow A C=B C$, from the diagram we know, $\overrightarrow{P C}=2 \overrightarrow{P M} \Rightarrow \lambda=-1$.