Let $\alpha, \beta ; \alpha>\beta$, be the roots of the equation $x^2-\sqrt{2} x-\sqrt{3}=0$. Let $P_n=\alpha^n-\beta^n, n \in N$. Then $(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$ is equal to :

  • [JEE MAIN 2024]
  • A

     $10 \sqrt{2} \mathrm{P}_9$

  • B

     $10 \sqrt{3} \mathrm{P}_9$

  • C

     $11 \sqrt{2} \mathrm{P}_9$

  • D

     $11 \sqrt{3} \mathrm{P}_9$

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  • [JEE MAIN 2021]