A) \[\frac{9}{4}\]
B) \[\frac{4}{9}\]
C) \[1\]
D) \[3\sqrt{3}\]
Correct Answer: A
Solution :
Given, in\[\Delta ABC\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[1({{c}^{2}}-ab)-a(c-a)+b(b-c)=0\] \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca=0\] \[\Rightarrow \]\[2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca=0\] \[+({{c}^{2}}+{{a}^{2}}-2ca)=0\] \[\Rightarrow \]\[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=0\] Here, sum of squares of three members can be zero if and only if\[a=b=c\] \[\Rightarrow \]\[\Delta ABC\]is equilateral. \[\Rightarrow \]\[\angle A=\angle B=\angle C={{60}^{o}}\] \[\therefore \]\[{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C\] \[=({{\sin }^{2}}{{60}^{o}}+{{\sin }^{2}}{{60}^{o}}+{{\sin }^{2}}{{60}^{o}})\] \[=3\times {{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=\frac{9}{4}\]
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