Let $a,b,c$ be such that $(b+a)c\ne 0$. If
$$\left|\begin{array}{ccc}a& a+1& a-1\\ b& b+1& b-1\\ c& c-1& c+1\end{array}\right|+\left|\begin{array}{ccc}a+1& b+1& c-1\\ a-1& b-1& c+1\\ (-1{)}^{n}a& (-1{)}^{n}b& (-1{)}^{n+1}c\end{array}\right|=0$$
then the value of '$n$ ' is:

Let the first determinant be ${\Delta }_1$.By swapping rows and columns (transpose) and performing row transformations on the second determinant, it can be seen that the equation holds zero only when the sign of the rows matches appropriately.Specifically, if $n$ is an odd integer, the terms in the last row of the second determinant will be $-a,-b,c$.Through row and column swaps, the second determinant becomes the negative of the first only if $n$ is odd.