A) abc
B) \[a+b+c\]
C) \[ab+bc+ca\]
D) None of these
Correct Answer: D
Solution :
\[\left| \,\begin{matrix} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b+c-a & c+a-b & a+b-c \\ \end{matrix}\, \right|\] = \[\left| \,\begin{matrix} 0 & 0 & 1 \\ b-a & c-b & a+b \\ 2\,(b-a) & 2\,(c-b) & a+b-c \\ \end{matrix}\, \right|\,=0\].
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