If the points $\mathrm{P}$ and $\mathrm{Q}$ are respectively the circumcenter and the orthocentre of a $\triangle \mathrm{ABC}$, then $\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$ is equal to :

Question

$\overrightarrow {QP} $

$\overrightarrow {PQ} $

$2\overrightarrow {PQ} $

$2\overrightarrow {QP} $

Solution

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