Let A, B, C be three angles such that sin A + | vedclass

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(C) Given $\sin A + \sin B + \sin C = 0$. We know the identity $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$. Thus,$\sin 3A + \sin 3B + \sin 3C = 3(\

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$-\frac{1}{12}$

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(C) Given $\sin A + \sin B + \sin C = 0$. We know the identity $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$. Thus,$\sin 3A + \sin 3B + \sin 3C = 3(\sin A + \sin B + \sin C) - 4(\sin^3 A + \sin^3 B + \sin^3 C)$. Since $\sin A + \sin B + \sin C = 0$,we have $\sin 3A + \sin 3B + \sin 3C = -4(\sin^3 A + \sin^3 B + \sin^3 C)$. Using the identity for $x+y+z=0 \implies x^3+y^3+z^3 = 3xyz$,we have $\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C$. Substituting this,$\sin 3A + \sin 3B + \sin 3C = -4(3 \sin A \sin B \sin C) = -12 \sin A \sin B \sin C$. Therefore,$\frac{\sin A \sin B \sin C}{\sin 3A + \sin 3B + \sin 3C} = \frac{\sin A \sin B \sin C}{-12 \sin A \sin B \sin C} = -\frac{1}{12}$.

Let $A, B, C$ be three angles such that $\sin A + \sin B + \sin C = 0$. Then,the value of $\frac{\sin A \sin B \sin C}{\sin 3A + \sin 3B + \sin 3C}$ (wherever defined) is:

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