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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=OQOR+OPOS
Then the triangle PQR has S as its

A
Circumcentre
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B
Orthocenter
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C
Incentre
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D
Centroid
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Solution

The correct option is C Orthocenter
Given OPOQ+OROS=OROP+OQOS
pq+rs=rp+qs
p(qr)=s(qr)
(ps)(qr)=0
SPQR
Similarly
QSPR
RSPQ
So S is orthocentre.

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