If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$, is $\alpha \times 2^8 \times \sqrt[5]{3}$, then $25 \alpha$ is equal to :

  • [JEE MAIN 2024]
  • A

    $639$

  • B

    $724$

  • C

    $693$

  • D

    $742$

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