If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ cuts the coordinate axes at points $A, B, C$ respectively,then the area of the triangle $ABC$ is

  • A
    $\sqrt{14}$ sq. units
  • B
    $3 \sqrt{14}$ sq. units
  • C
    $\frac{1}{\sqrt{14}}$ sq. units
  • D
    $3 \sqrt{13}$ sq. units

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