Let $A, B, C$ are three angles such that $sinA + sinB + sinC = 0,$ then

$ \frac {sinAsin BsinC}{(sin 3A+ sin 3B+ sin 3C)}$ (wherever definied) is -

  • A

    $12$

  • B

    $-12$

  • C

    $  - \frac{1}{12}$

  • D

    $\frac{1}{12}$

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  • [IIT 1979]