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(A) Given the determinant $\left| \begin{array}{ccc} x + 1 & x + 2 & x + 3 \ x + 2 & x + 3 & x + 4 \ x + a & x + b & x + c \end{array} ight| = 0$.

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$A.P.$

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(A) Given the determinant $\left| \begin{array}{ccc} x + 1 & x + 2 & x + 3 \ x + 2 & x + 3 & x + 4 \ x + a & x + b & x + c \end{array} ight| = 0$. Applying column operations $C_1 	o C_1 - C_2$ and $C_2 	o C_2 - C_3$: $\left| \begin{array}{ccc} -1 & -1 & x + 3 \ -1 & -1 & x + 4 \ a - b & b - c & x + c \end{array} ight| = 0$. Applying row operation $R_1 	o R_1 - R_2$: $\left| \begin{array}{ccc} 0 & 0 & -1 \ -1 & -1 & x + 4 \ a - b & b - c & x + c \end{array} ight| = 0$. Expanding along the first row: $(-1) \cdot [(-1)(b - c) - (-1)(a - b)] = 0$ $(-1) \cdot [-b + c + a - b] = 0$ $2b - a - c = 0 \implies a + c = 2b$. This condition implies that $a, b, c$ are in $A.P.$ Alternatively,setting $x = 0$: $\left| \begin{array}{ccc} 1 & 2 & 3 \ 2 & 3 & 4 \ a & b & c \end{array} ight| = 0$ $1(3c - 4b) - 2(2c - 4a) + 3(2b - 3a) = 0$ $3c - 4b - 4c + 8a + 6b - 9a = 0$ $-a + 2b - c = 0 \implies 2b = a + c$. Thus,$a, b, c$ are in $A.P.$

If $\left| \begin{array}{ccc} x + 1 & x + 2 & x + 3 \\ x + 2 & x + 3 & x + 4 \\ x + a & x + b & x + c \end{array} \right| = 0$,then $a, b, c$ are in

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