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

  • A
    $2 \overrightarrow{QP}$
  • B
    $\overrightarrow{QP}$
  • C
    $2 \overrightarrow{PQ}$
  • D
    $\overrightarrow{PQ}$

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