Let \(O\) be the origin and let \(P Q R\) be an arbitrary triangle. The point \(S\) \(\overline{O P} \cdot \overline{O Q}+\overline{O R} \cdot \overline{O S}=\overline{O R} \cdot \overline{O P}+\overline{O Q} \cdot \overline{O S}=\) \(\overline{O Q} \overline{O Q} \cdot \overline{O R}+\overline{O P} \cdot \overline{O S}\) that \(\overline{O P} \cdot \overline{O Q}+\overline{O R} \cdot \overline{O S}=\overline{O R} \cdot \overline{O P}+\overline{O Q} \cdot \overline{O S}=\) \(\overline{O Q} \cdot \overline{O R}+\overline{O P} \cdot \overline{O S}\), then the triangle \(P Q R\) has \(S\) as its

\(\overrightarrow{O P} \cdot \overrightarrow{O Q}+\overrightarrow{O R} \cdot \overrightarrow{O S}=\overrightarrow{O R} \cdot \overrightarrow{O P}+\overrightarrow{O Q} \cdot \overrightarrow{O S} \)
\( \Rightarrow \overrightarrow{O P} \cdot(\overrightarrow{O Q}-\overrightarrow{O R})=\overrightarrow{O S} \cdot(\overrightarrow{O Q}-\overrightarrow{O R}) \)
\( \Rightarrow \overrightarrow{O P} \cdot \overrightarrow{R Q}=\overrightarrow{O S} \cdot \overrightarrow{R Q} \)
\( \Rightarrow \overrightarrow{R Q} \cdot(\overrightarrow{O P}-\overrightarrow{O S})=0 \)
\( \Rightarrow \overrightarrow{R Q} \cdot \overrightarrow{P S}=0 \)
\( \Rightarrow \overrightarrow{P S} \perp \overrightarrow{Q R}\)
Similarly \(\overrightarrow{Q S} \perp \overrightarrow{P R}\) and \(\overrightarrow{R S} \perp \overrightarrow{P Q}\)
i.e., \(S\) is the orthocenter