A) \[x-(a+b+c)\]
B) \[9{{x}^{2}}+a+b+c\]
C) \[a+b+c\]
D) 0
Correct Answer: D
Solution :
Let \[A=\left| \,\begin{matrix} x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c \\ \end{matrix}\, \right|\] Applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}},\] we get, Þ \[A=\,\left| \,\begin{matrix} x+2 & 1 & x+a \\ x+4 & 1 & x+b \\ x+6 & 1 & x+c \\ \end{matrix}\, \right|\] Applying \[\left| \,\begin{matrix} a & {{a}^{3}} & {{a}^{4}} \\ b & {{b}^{3}} & {{b}^{4}} \\ c & {{c}^{3}} & {{c}^{4}} \\ \end{matrix}\, \right|+\left| \,\begin{matrix} a & {{a}^{3}} & -1 \\ b & {{b}^{3}} & -1 \\ c & {{c}^{3}} & -1 \\ \end{matrix}\, \right|=0\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] Þ \[A=\left| \,\begin{matrix} x+2 & 1 & x+a \\ 2 & 0 & b-a \\ 4 & 0 & c-a \\ \end{matrix}\, \right|\,=\,-1\,(2c-2a-4b+4a)\] = \[2(2b-c-a)\] \[\because \] a, b, c are in A.P. Þ A = 0.
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