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(C) Let $D_1 = \begin{vmatrix} a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1 \end{vmatrix}$ and $D_2 = \begin{vmatrix} a+1 & b+1 & c-1 \ a-1 & b-1

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Any odd integer

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(C) Let $D_1 = \begin{vmatrix} a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1 \end{vmatrix}$ and $D_2 = \begin{vmatrix} a+1 & b+1 & c-1 \ a-1 & b-1 & c+1 \ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{vmatrix}$.We are given $D_1 + D_2 = 0$.In $D_2$,take $(-1)^n$ common from the third row: $D_2 = (-1)^n \begin{vmatrix} a+1 & b+1 & c-1 \ a-1 & b-1 & c+1 \ a & -b & c \end{vmatrix}$.Now,perform column operations on $D_2$: Swap $C_1$ and $C_2$,then swap $C_2$ and $C_3$. Two swaps result in no change in sign.$D_2 = (-1)^n \begin{vmatrix} a & a+1 & a-1 \ -b & b+1 & b-1 \ c & c-1 & c+1 \end{vmatrix} = (-1)^n D_1$.Thus,$D_1 + (-1)^n D_1 = 0 \Rightarrow (1 + (-1)^n) D_1 = 0$.For this to hold for arbitrary $a, b, c$,we must have $1 + (-1)^n = 0$,which implies $(-1)^n = -1$.This is true when $n$ is an odd integer.

Let $a, b, c$ be such that $(b+c) \neq 0$ and $\left|\begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array}\right|+\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{array}\right|=0$. Then the value of $n$ is

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