A) 0
B) 1
C) 2
D) \[3abc\]
Correct Answer: A
Solution :
We have \[2\,\,\left| \,\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{2}}-bc & {{b}^{2}}-ac & {{c}^{2}}-ab \\ \end{matrix}\, \right|\] = \[2\,\left| \,\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ \end{matrix}\, \right|-2\left| \,\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ bc & ac & ab \\ \end{matrix}\, \right|\] = \[2\,\left| \,\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ \end{matrix}\, \right|-\frac{2}{abc}\left| \,\begin{matrix} a & b & c \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ abc & abc & abc \\ \end{matrix}\, \right|\] Applying \[{{C}_{1}}(a),{{C}_{2}}(b),{{C}_{3}}(c)\] \[=2\,\left| \,\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ \end{matrix}\, \right|-\frac{2}{abc}(abc)\,\left| \,\begin{matrix} a & b & c \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ 1 & 1 & 1 \\ \end{matrix}\, \right|=0\].
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