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