Let $O$ be the origin and let $P Q R$ be an arbitrary triangle. The point $S$ is such that \[ \overrightarrow{O P} \cdot \overrightarrow{O Q}+\overrightarrow{O R} \cdot \overrightarrow{O S}=\overrightarrow{O R} \cdot \overrightarrow{O P}+\overrightarrow{O Q} \cdot \overrightarrow{O S}=\overrightarrow{O Q} \cdot \overrightarrow{O R}+\overrightarrow{O P} \cdot \overrightarrow{O S} \] Then the triangle $P Q R$ has $S$ as its