A) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1\]
B) \[bc+ca+ab=0\]
C) \[a+b+c=0\]
D) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=bc+ca+ab\]
Correct Answer: B
Solution :
Accordingly, \[\left| \,\begin{matrix} -bc & {{b}^{2}}+bc & {{c}^{2}}+bc \\ {{a}^{2}}+ac & -ac & {{c}^{2}}+ac \\ {{a}^{2}}+ab & {{b}^{2}}+ab & -ab \\ \end{matrix}\, \right|=0\] \[\Rightarrow {{(ab+bc+ca)}^{3}}=0\Rightarrow ab+bc+ca=0.\]You need to login to perform this action.
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