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(B) Given determinant is $\Delta = \left| {\begin{array}{ccc} 1 & 1+ac & 1+bc \ 1 & 1+ad & 1+bd \ 1 & 1+ae & 1+be \end{array}} ight|$.Applying the

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$0$

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(B) Given determinant is $\Delta = \left| {\begin{array}{ccc} 1 & 1+ac & 1+bc \ 1 & 1+ad & 1+bd \ 1 & 1+ae & 1+be \end{array}} ight|$.Applying the column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\Delta = \left| {\begin{array}{ccc} 1 & ac & bc \ 1 & ad & bd \ 1 & ae & be \end{array}} ight|$.Now,taking $a$ common from $C_2$ and $b$ common from $C_3$:$\Delta = ab \left| {\begin{array}{ccc} 1 & c & c \ 1 & d & d \ 1 & e & e \end{array}} ight|$.Since column $C_2$ and column $C_3$ are identical,the value of the determinant is $0$.

$\left| {\begin{array}{ccc} 1 & 1+ac & 1+bc \\ 1 & 1+ad & 1+bd \\ 1 & 1+ae & 1+be \end{array}} \right| = $

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