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(C) The area of the triangle is given by $\Delta = \frac{1}{2} ab \sin C$. Substituting the given values,$15 \sqrt{3} = \frac{1}{2} 	imes 6 	imes 10

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$3$

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(C) The area of the triangle is given by $\Delta = \frac{1}{2} ab \sin C$. Substituting the given values,$15 \sqrt{3} = \frac{1}{2} 	imes 6 	imes 10 	imes \sin C$,which simplifies to $\sin C = \frac{\sqrt{3}}{2}$. Since $\angle ACB$ is obtuse,$C = 120^{\circ}$. Using the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C = 6^2 + 10^2 - 2(6)(10) \cos 120^{\circ} = 36 + 100 - 120(-0.5) = 136 + 60 = 196$,so $c = 14$. The semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+10+14}{2} = 15$. The inradius $r = \frac{\Delta}{s} = \frac{15 \sqrt{3}}{15} = \sqrt{3}$. Therefore,$r^2 = (\sqrt{3})^2 = 3$.

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