If the points P and Q are respectively the circumcenter and the orthocenter of a $\mathrm{△}\mathrm{ABC}$, then $\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$ is

Let p be the origin 
$\begin{array}{l}\overrightarrow{\mathrm{PD}}=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}\Rightarrow 2\overrightarrow{\mathrm{PD}}=\overline{\mathrm{b}}+\overline{\mathrm{c}}\\ \overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}=\overline{\mathrm{b}}+\overline{\mathrm{c}}\\ \overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}=\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}\\ \overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}=3\overline{PG}\\ \overrightarrow{ and \mathrm{PQ}}=3\overline{PG}=\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}( \sin ce centroid divides Q,P in the ratio 2:1)\\ \overrightarrow{\therefore \mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}=\overrightarrow{\mathrm{PQ}}\end{array}$