$\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}-\frac{6}{\sqrt{8}+\sqrt{12}}=?$

  • A
    $1$
  • B
    $-\sqrt{3}$
  • C
    $\sqrt{3}+\sqrt{2}$
  • D
    $\sqrt{3}-\sqrt{2}$

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