Answer:
We have, \[\Rightarrow \] \[[applying\,\,{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\,\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}]\] \[\Rightarrow \] [expanding along \[{{C}_{1}}\]] \[\Rightarrow \] \[(c-a)(c-b)-(a-b)(b-a)=0\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca=0\] \[\Rightarrow \] \[2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca=0\] \[\Rightarrow \] \[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=0\] \[\Rightarrow \] \[a-b=0,\] \[b-c=0,\] \[c-a=0\] \[\Rightarrow \] a = b = c \[\Rightarrow \] \[\Delta ABC\] is an equilateral triangle. \[\Rightarrow \] \[A=B=C=\frac{\pi }{3}\] \[\therefore \] \[{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=3{{\sin }^{2}}\frac{\pi }{3}\] \[=3\times {{\left( \frac{\sqrt{3}}{2} \right)}^{2}}\] \[=\frac{9}{4}\]
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