If we know three vertices a, B, C of ⊿ ABC and a point P in the plane, satisfying PA + Pb + PC = 0, then point P is () A. Center of gravity

If we know three vertices a, B, C of ⊿ ABC and a point P in the plane, satisfying PA + Pb + PC = 0, then point P is () A. Center of gravity

Choose C, and point P is the center of gravity of △ ABC
The reasons are as follows.
Take the midpoint m of AB, connect PM and extend it to Q such that MQ = PM, then:
The quadrilateral apbq is a parallelogram
Thus, PA + Pb = PQ = 2pm
If PA + Pb + PC = 0, 2pm + PC = 0, i.e. points c, P and m are in a straight line and | PC | = 2 | PM |, then point P is a trisection point close to m of the central line cm of △ ABC, and point P is the center of gravity of △ ABC