Let the plane $P : 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$. If the intercept of $P$ on the $y$-axis is $1$,then the distance between $P$ and $L$ is:

  • A
    $\sqrt{14}$
  • B
    $\frac{6}{\sqrt{14}}$
  • C
    $\sqrt{\frac{2}{7}}$
  • D
    $\sqrt{\frac{7}{2}}$

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