A) - 1
B) 1
C) 2
D) -2
Correct Answer: C
Solution :
Operate \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}},\,{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] and take out \[a+b+c\] from \[{{C}_{2}}\] as well as from \[{{C}_{3}}\] to get \[\Delta ={{(a+b+c)}^{2}}\] \[\left| \,\begin{matrix} {{(b+c)}^{2}} & a-b-c & a-b-c \\ {{b}^{2}} & c+a-b & 0 \\ {{c}^{2}} & 0 & a+b-c \\ \end{matrix}\, \right|\] (Operate\[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}-{{R}_{3}}\]) = \[{{(a+b+c)}^{2}}\,\left| \,\begin{matrix} 2bc & -2c & -2b \\ {{b}^{2}} & c+a-b & 0 \\ {{c}^{2}} & 0 & a+b-c \\ \end{matrix}\, \right|\] (Operate \[{{C}_{2}}\to {{C}_{2}}+\frac{1}{b}{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}+\frac{1}{c}{{C}_{1}})\] = \[{{(a+b+c)}^{2}}\left| \begin{matrix} 2bc & 0 & 0 \\ {{b}^{2}} & c+a & \frac{{{b}^{2}}}{c} \\ {{c}^{2}} & \frac{{{c}^{2}}}{b} & a+b \\ \end{matrix} \right|\] \[=\,{{(a+b+c)}^{2}}[2bc\{(a+b)\,(c+a)-bc\}]\]\[=\,2abc{{(a+b+c)}^{3}}\].
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