In $\triangle ABC$,the points $P, Q, R$ divide $BC, CA, AB$ in the ratio $3:4, 2:5, 9:5$,respectively,and the point $D$ divides $BC$ in the ratio $2:3$. If $\vec{AP} + \vec{BQ} + \vec{CR} = k \vec{AD}$,then $(14k + 1) : (14k - 1) = $

  • A
    $7:5$
  • B
    $4:3$
  • C
    $4:5$
  • D
    $3:2$

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