Let O be the origin and let PQR be an arbitrary triangle. The point S is such that
⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶
OP . OQ + OR . OS = OR . OP + OQ . OS = OQ . OR + OP . OS
Then the triangle PQR has S as its
[A] centroid [B] circumcentre
[C] incentre [D] orthocenter
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centroid is the correct answer ..........
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