If $P$ is a point equidistant from all the vertices $A(-1, 3)$,$B(3, 5)$,and $C(5, 7)$ of a triangle $ABC$,then $PA=$

  • A
    $11$
  • B
    $\sqrt{140}$
  • C
    $13$
  • D
    $\sqrt{130}$

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