If $\alpha$ satisfies the equation $x^2+x+1=0$ and $(1+\alpha)^7=A+B\alpha+C\alpha^2$,where $A, B, C \geq 0$,then $5(3A-2B-C)$ is equal to:

  • A
    $6$
  • B
    $5$
  • C
    $7$
  • D
    $3$

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