JEE Advanced 2017 Paper 2 Mathematics Q2

Mathematics

Problem

Let OO be the origin and let PQRPQR be an arbitrary triangle. The point SS is such that OP→⋅OQ→+OR→⋅OS→=OR→⋅OP→+OQ→⋅OS→=OQ→⋅OR→+OP→⋅OS→\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}. Then the triangle PQRPQR has SS as its

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