A) 1
B) 0
C) 3
D) \[a+b+c\]
Correct Answer: B
Solution :
Applying \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]and \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\], we get \[\left| \,\begin{matrix} 1 & ac & bc \\ 1 & ad & bd \\ 1 & ae & be \\ \end{matrix}\, \right|=ab\left| \,\begin{matrix} 1 & c & c \\ 1 & d & d \\ 1 & e & e \\ \end{matrix}\, \right|=0\], \[\{\because {{C}_{2}}\equiv {{C}_{3}}\}\].
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