$(a)\;2\qquad(b)\;4\qquad(c)\;3\qquad(d)\;1$

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Given :

$a=6$cm

Area $\Delta=12cm^2$

$B-C=60^{\large\circ}$

Now $\Delta=\large\frac{1}{2}$$ab\sin C$

$\qquad\;\;\;=\large\frac{1}{2}$$a.k\sin B\sin C$

$\qquad\;\;\;=\large\frac{1}{2}$$a.\large\frac{a}{\sin A}$$\sin B\sin C$

$\large\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$=k$

$\Delta=\large\frac{1}{2}\frac{a^2\sin B\sin C}{\sin A}$

$12=\large\frac{1}{2}\frac{36\sin B\sin C}{\sin A}$

$\large\frac{2}{3}=\frac{2\sin B\sin C}{2\sin A}$

$\large\frac{2}{3}=\frac{\cos(B-C)-\cos(B+C)}{2\sin A}$

$\large\frac{2}{3}=\frac{\cos (60^{\large\circ})-\cos(\pi-A)}{2\sin A}$

$\large\frac{2}{3}=\frac{1+2\cos A}{4\sin A}$

$4\sin A=3+6\cos A$

$8\sin A-6\cos A=3$

Hence (c) is the correct answer.

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