$\left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|+\left|\begin{array}{ccc}a+1& b+1& c-1\\ a-1& b-1& c+1\\ {\left(-1\right)}^{n+2}a& {\left(-1\right)}^{n-1}b& {\left(-1\right)}^{n}c\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|+{\left(-1\right)}^n\left|\begin{array}{ccc}a+1& b+1& c-1\\ a-1& b-1& c+1\\ a& -b& c\end{array}\right|=0$$$

$\Rightarrow \left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|+{\left(-1\right)}^n\left|\begin{array}{ccc}a+1& a-1& a\\ b+1& b-1& -b\\ c-1& c+1& c\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|+{\left(-1\right)}^n\left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|=0$

$\Rightarrow (1+{\left(-1\right)}^n)\left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|=0$

Hence, $n$ is an odd integer.