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(D) Let $\Delta = \left| \begin{array}{ccc} x+2 & x+3 & x+a \ x+4 & x+5 & x+b \ x+6 & x+7 & x+c \end{array} ight|$.Applying the column operation $

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(D) Let $\Delta = \left| \begin{array}{ccc} x+2 & x+3 & x+a \ x+4 & x+5 & x+b \ x+6 & x+7 & x+c \end{array} ight|$.Applying the column operation $C_2 	o C_2 - C_1$,we get:$\Delta = \left| \begin{array}{ccc} x+2 & 1 & x+a \ x+4 & 1 & x+b \ x+6 & 1 & x+c \end{array} ight|$.Now,applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$,we get:$\Delta = \left| \begin{array}{ccc} x+2 & 1 & x+a \ 2 & 0 & b-a \ 4 & 0 & c-a \end{array} ight|$.Expanding along the second column:$\Delta = -1 	imes [2(c-a) - 4(b-a)]$$\Delta = -1 	imes [2c - 2a - 4b + 4a]$$\Delta = -1 	imes [2c + 2a - 4b]$$\Delta = 2(2b - a - c)$.Since $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $2b - a - c = 0$.Therefore,$\Delta = 2(0) = 0$.

If $a, b, c$ are in $A.P.$,then the value of $\left| \begin{array}{ccc} x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c \end{array} \right|$ is

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