Let $a, b, c$ be such that $b (a + c) ≠ 0$. 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

any odd integer

We have,

$\begin{vmatrix}a &a +1 &a-1\\-b &b+1 &b-1\\c &c-1 &c+1\end{vmatrix}+\begin{vmatrix}a+1&b+1&c-1\\a-1&b-1&c+1\\(-1)^{n+2} a &(-1)^{n+1}b &(-1)^n c\end{vmatrix}=0$

$⇒\begin{vmatrix}a &-b&c\\a+1 &b+1 &c-1\\a-1& b-1 &c+1\end{vmatrix}+(-1)^n\begin{vmatrix}a+1 &b+1 &c-1\\a-1& b-1 &c+1\\a &-b&c\end{vmatrix}=0$ [Using $|A| =|A|^T$ in first determinant]

$⇒\begin{vmatrix}a &-b&c\\a+1 &b+1 &c-1\\a-1& b-1 &c+1\end{vmatrix}+(-1)^{n+2}\begin{vmatrix}a &-b&c\\a+1 &b+1 &c-1\\a-1& b-1 &c+1\end{vmatrix}=0$ [Applying first $R_1 ↔R_3$ and then $R_2 ↔ R_3$]

$⇒\begin{vmatrix}a &-b&c\\a+1 &b+1 &c-1\\a-1& b-1 &c+1\end{vmatrix}\{1+(-1)^{n+2}\}=0$

$⇒\begin{vmatrix}a &-b&c\\1 &2b+1 &-1\\-1& 2b-1 &1\end{vmatrix}\{1+(-1)^{n+2}\}=0$ [Applying $R_2 → R_2-R_1, R_3 → R_3 - R_1$]

$⇒\begin{vmatrix}a &-b&c\\1 &2b+1 &-1\\0& 4b &0\end{vmatrix}\{1+(-1)^{n+2}\}=0$ [Applying $R_3 →R_3 + R_2$]

$⇒b (a + c)\{1+(-1)^{n+2}\}=0$

$⇒1+(-1)^{n+2}=0$ ⇒ n is any odd integer