Let O be the origin and PQR be an arbitrary t | vedclass

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(C) Given the condition: $\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \o

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Orthocentre.

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(C) Given the condition: $\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS} = \overrightarrow{OQ} \cdot \overrightarrow{OR} + \overrightarrow{OP} \cdot \overrightarrow{OS}$.Consider the first equality: $\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS}$.Rearranging the terms: $\overrightarrow{OP} \cdot \overrightarrow{OQ} - \overrightarrow{OR} \cdot \overrightarrow{OP} = \overrightarrow{OQ} \cdot \overrightarrow{OS} - \overrightarrow{OR} \cdot \overrightarrow{OS}$.Factoring out the common vectors: $\overrightarrow{OP} \cdot (\overrightarrow{OQ} - \overrightarrow{OR}) = \overrightarrow{OS} \cdot (\overrightarrow{OQ} - \overrightarrow{OR})$.This implies: $(\overrightarrow{OP} - \overrightarrow{OS}) \cdot (\overrightarrow{OQ} - \overrightarrow{OR}) = 0$.Since $\overrightarrow{OP} - \overrightarrow{OS} = \overrightarrow{SP}$ and $\overrightarrow{OQ} - \overrightarrow{OR} = \overrightarrow{RQ}$,we have $\overrightarrow{SP} \cdot \overrightarrow{RQ} = 0$,which means $\overrightarrow{PS} \perp \overrightarrow{QR}$.Similarly,by considering the other parts of the equality,we get $\overrightarrow{QS} \perp \overrightarrow{PR}$ and $\overrightarrow{RS} \perp \overrightarrow{PQ}$.Since $S$ is the point where the altitudes of the triangle $PQR$ intersect,$S$ is the orthocentre.

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