5. If $P$ is the circumcenter of $\triangle A B C$, and
$$
\overrightarrow{P A}+\overrightarrow{P B}+\lambda \overrightarrow{P C}=\mathbf{0}, \angle C=120^{\circ} \text {. }
$$

then the value of the real number $\lambda$ is $\qquad$
5. -1 .

Let the circumradius of $\triangle ABC$ be $R$.
From the given information, we have
$$
\begin{array}{l}
|\overrightarrow{P A}+\overrightarrow{P B}|^{2}=\lambda^{2}|\overrightarrow{P C}|^{2} \\
=|\overrightarrow{P A}|^{2}+|\overrightarrow{P B}|^{2}-2|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{C}{2} \\
=R^{2}+R^{2}-2 R^{2} \cdot \frac{1}{2}=R^{2}=\lambda^{2} R^{2} \\
\Rightarrow \lambda= \pm 1 .
\end{array}
$$

From the problem, we get $\lambda=-1$.