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

Solution :
$$\eqalign{ & \overrightarrow {OP} .\overrightarrow {OQ} + \overrightarrow {OR} .\overrightarrow {OS} = \overrightarrow {OR} .\overrightarrow {OP} + \overrightarrow {OQ} .\overrightarrow {OS} \cr & \Rightarrow \left( {\overrightarrow {OQ} - \overrightarrow {OR} } \right).\overrightarrow {OP} - \left( {\overrightarrow {OQ} - \overrightarrow {OR} } \right).\overrightarrow {OS} = 0 \cr & \Rightarrow \left( {\overrightarrow {OQ} - \overrightarrow {OR} } \right).\left( {\overrightarrow {OP} - \overrightarrow {OS} } \right) = 0 \cr & \Rightarrow \overrightarrow {RQ} .\overrightarrow {SP} = 0 \cr & \Rightarrow RQ\, \bot \,SP.....(1) \cr & {\text{Also }}\overrightarrow {OR} .\overrightarrow {OP} + \overrightarrow {OQ} .\overrightarrow {OS} = \overrightarrow {OQ} .\overrightarrow {OR} + \overrightarrow {OP} .\overrightarrow {OS} \cr & \Rightarrow \overrightarrow {OR} .\left( {\overrightarrow {OP} - \overrightarrow {OQ} } \right) - \overrightarrow {OS} .\left( {\overrightarrow {OP} - \overrightarrow {OQ} } \right) = 0 \cr & \Rightarrow \left( {\overrightarrow {OP} - \overrightarrow {OQ} } \right).\left( {\overrightarrow {OR} - \overrightarrow {OS} } \right) = 0 \cr & \Rightarrow \overrightarrow {QP} .\overrightarrow {SR} = 0 \cr & \Rightarrow QP\, \bot \,SR.....(2) \cr} $$
From (1) and (2) $$S$$ should be the orthocenter of $$\Delta PQR.$$