A) \[\frac{4}{9}\]
B) \[\frac{9}{4}\]
C) \[3\sqrt{3}\]
D) 1
Correct Answer: B
Solution :
Here we have \[\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[{{c}^{2}}-ab-a\,(c-a)+b\,(b-c)=0\] \[\Rightarrow \] \[{{c}^{2}}-ab-ac+{{a}^{2}}+{{b}^{2}}-bc=0\] \[\Rightarrow \] \[{{c}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca=0\] \[\Rightarrow \] \[\frac{1}{2}[2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca]=0\] \[\Rightarrow \] \[\frac{1}{2}[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]=0\] So, \[\Delta \,ABC\] is equilateral \[\therefore \] \[\angle \,A={{60}^{o}},\,\angle \,B={{60}^{o}},\,\angle C={{60}^{o}}\] \[{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C\] \[={{\sin }^{2}}{{60}^{o}}+{{\sin }^{2}}{{60}^{o}}+{{\sin }^{2}}{{60}^{o}}\] \[={{\left( \frac{\sqrt{3}}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}\] \[=3\times \frac{3}{4}=\frac{9}{4}\]
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