A) \[a=b\ne c\]
B) \[a\ne b=c\]
C) \[a\ne b\ne c\]
D) \[a=b=c\]
Correct Answer: D
Solution :
\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca=0\] \[\Rightarrow \]\[2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca=0\] \[\Rightarrow \]\[({{a}^{2}}-2ab+{{b}^{2}})({{b}^{2}}-2bc+{{c}^{2}})\]\[+({{c}^{2}}-2ac+{{a}^{2}})=0\] \[\Rightarrow \]\[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=0\] \[\Rightarrow \]\[a-b=0,\,\,b-c=0,\,\,c-a=0\] \[\Rightarrow \]\[a=b,\,\,b=c\]and\[c=a\] \[\Rightarrow \]\[a=b=c\]
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