3. Given that $P$ is a point on the plane of $\triangle A B C$, satisfying $\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}=2 \overrightarrow{A B}$. If $S_{\triangle A B C}=6$, then $S_{\triangle P A B}$ $=(\quad)$.
(A) 4
(B) 3
(C) 2
(D) 1
3. C.

From $\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}=2 \overrightarrow{A B}=2(\overrightarrow{P B}-\overrightarrow{P A})$, we get $3 \overrightarrow{P A}=\overrightarrow{P B}-\overrightarrow{P C}=\overrightarrow{C B}$.
Therefore, $S_{\triangle P A B}=\frac{1}{3} S_{\triangle A B C}=2$.