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Question

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that
OPOQ+OROS=OROP+OQOS=OQ.OR+OP.OS
Then the triangle PQR has S as its

A
centroid
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B
circumcentre
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C
incentre
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D
orthocenter
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Solution

The correct option is D orthocenter
OPOQ+OROS=OROP+OQOSOP(OQOR)=OS(OQOR)(OPOS)(OQOR)=0SPRQ=0SPRQ
Similarly, SRQP and SQPR
Hence, S is the orthocentre of triangle PQR

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