Let O be the origin and let PQR be an arbitrary triangle. The point S is such that $\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OS}} =\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OS}}=\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OR}}+\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OS}}$ Then the triangle PQR has S as its