• # question_answer In$\Delta \,\,ABC$, if$\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \\ \end{matrix} \right|=0$, then${{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=$ A) $\frac{9}{4}$                        B) $\frac{4}{9}$C) $1$                             D) $3\sqrt{3}$

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$ $=({{\sin }^{2}}{{60}^{o}}+{{\sin }^{2}}{{60}^{o}}+{{\sin }^{2}}{{60}^{o}})$ $=3\times {{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=\frac{9}{4}$