The distance of the point $(2, 4, 0)$ from the point of intersection of the lines $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}$ and $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}$ is

  • A
    $3$ units
  • B
    $3 \sqrt{3}$ units
  • C
    $2$ units
  • D
    $2 \sqrt{3}$ units

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