A) \[abc\]
B) \[2abc\]
C) \[3abc\]
D) \[4abc\]
Correct Answer: D
Solution :
\[\left| \,\begin{matrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \\ \end{matrix}\, \right|\,=\,\left| \,\begin{matrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \\ \end{matrix}\, \right|\] {by \[{{R}_{1}}\to {{R}_{1}}-({{R}_{2}}+{{R}_{3}})\}\] = \[2c.b(a+b-c)\,-2b.c(b-c-a)\,=4abc\].
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