Let $$a, b, c$$ be such that $$b(a + c) \neq 0$$. If $$$\begin{vmatrix} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{vmatrix} + \begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^n c \end{vmatrix} = 0,$$$ then the value of '$$n$$' is

Let the two determinants be

$$D_1=\begin{vmatrix} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{vmatrix},\qquad D_2=\begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{\,n+2}a & (-1)^{\,n+1}b & (-1)^{\,n}c \end{vmatrix}.$$

The given condition is $$D_1+D_2=0$$ with $$b(a+c)\neq 0.$$ We evaluate each determinant separately.

Case 1: Evaluation of $$D_1$$

Apply the column transformations $$C_2\to C_2-C_1,\; C_3\to C_3-C_1$$ (determinant remains unchanged):

$$D_1=\begin{vmatrix} a & 1 & -1 \\ -b & 2b+1 & 2b-1 \\ c & -1 & 1 \end{vmatrix}.$$

Expanding along the first row,

$$\begin{aligned} D_1 & = a\bigl((2b+1)(1)-(2b-1)(-1)\bigr) \\ &\quad -1\bigl((-b)(1)-(2b-1)c\bigr) \\ &\quad +(-1)\bigl((-b)(-1)-(2b+1)c\bigr). \end{aligned}$$

Simplifying each term,

$$D_1 = 4ab + (b+2bc-c) + (-b+2bc+c)=4ab+4bc.$$

Thus

$$D_1 = 4b(a+c).$$

Case 2: Evaluation of $$D_2$$

Write $$(-1)^{\,n}=s\;(=\pm1).$$ Then $$(-1)^{\,n+1}=-s,\; (-1)^{\,n+2}=s.$$ So

$$D_2=\begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ sa & -sb & sc \end{vmatrix}=s\, \begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ a & -b & c \end{vmatrix}.$$ Call the second determinant $$D'_2.$$

Compute $$D'_2$$ by the row operation $$R_1\to R_1-R_2$$:

$$D'_2=\begin{vmatrix} 2 & 2 & -2 \\ a-1 & b-1 & c+1 \\ a & -b & c \end{vmatrix} =2\begin{vmatrix} 1 & 1 & -1 \\ a-1 & b-1 & c+1 \\ a & -b & c \end{vmatrix}.$$

Expanding along the first row:

$$\begin{aligned} D'_2 &= 2\Bigl[(1)\bigl((b-1)c-(c+1)(-b)\bigr) \\ &\qquad -(1)\bigl((a-1)c-(c+1)a\bigr) \\ &\qquad +(-1)\bigl((a-1)(-b)-(b-1)a\bigr)\Bigr]. \end{aligned}$$

Simplifying inside the brackets gives $$2b(c+a).$$ Hence

$$D'_2 = 2\cdot 2b(a+c)=4b(a+c).$$

Therefore

$$D_2 = s\;D'_2 = (-1)^{\,n}\,4b(a+c).$$

Case 3: Applying the given relation

$$D_1+D_2=0\;\Longrightarrow\;4b(a+c)+4b(a+c)(-1)^{\,n}=0.$$

Since $$b(a+c)\neq 0,$$ divide both sides by $$4b(a+c)$$ to get

$$1+(-1)^{\,n}=0\;\Longrightarrow\;(-1)^{\,n}=-1.$$

This happens exactly when $$n$$ is an odd integer.

Hence the required value of $$n$$ is any odd integer.

Option C which is: any odd integer