A) \[0\]
B) \[abc\]
C) \[ab+bc+ca\]
D) \[abc\,(a+b+c)\]
Correct Answer: A
Solution :
[a] \[\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ac & b(c+a) \\ 1 & ab & c(a+b) \\ \end{matrix} \right|\] Applying \[{{C}_{3}}\to {{C}_{2}}+{{C}_{3}}\] \[\left| \begin{matrix} 1 & bc & ab+bc+ac \\ 1 & ac & ab+bc+ac \\ 1 & ab & ab+bc+ac \\ \end{matrix} \right|\] \[=(ab+bc+ac)\left| \begin{matrix} 1 & bc & 1 \\ 1 & ac & 1 \\ 1 & ab & 1 \\ \end{matrix} \right|\] \[=(ab+bc+ac)\times 0=0\]
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