If the plane $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$ cuts the coordinate axes at points $A, B$,and $C$,then the area of the triangle $ABC$ is

  • A
    $\frac{\sqrt{61}}{2}$ sq. units
  • B
    $2 \sqrt{61}$ sq. units
  • C
    $\sqrt{61}$ sq. units
  • D
    $3 \sqrt{61}$ sq. units

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