Given,

$A=\left[\begin{array}{lll}0& 1& 2\\ a& 0& 3\\ 1& c& 0\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{lll}0& 1& 2\\ a& 0& 3\\ 1& c& 0\end{array}\right]\left[\begin{array}{lll}0& 1& 2\\ a& 0& 3\\ 1& c& 0\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{ccc}a+2& 2c& 3\\ 3& a+3c& 2a\\ ca& 1& 2+3c\end{array}\right]$

Now finding,

$A^3=\left[\begin{array}{ccc}a+2& 2c& 3\\ 3& a+3c& 2a\\ ca& 1& 2+3c\end{array}\right]\left[\begin{array}{ccc}0& 1& 2\\ a& 0& 3\\ 1& c& 0\end{array}\right]$

$\Rightarrow A^3=\left[\begin{array}{ccc}2ca+3& \alpha +2+3c& 2a+4+6c\\ a^2+3ca+2a& 3+2ac& 6+3a+9c\\ a+2+3c& ca+2c+3c^2& 2ca+3\end{array}\right]$

Now equating $A^3=A$ and comparing both side we get,

$2ca+3=0\Rightarrow c=\frac{-3}{2a}$

And $a+2+3c=1$

$\Rightarrow a+2+3\left(\frac{-3}{2a}\right)=1$

$\Rightarrow a+1-\frac{9}{2a}=0$

$\Rightarrow 2a^2+2a-9=0$

$\Rightarrow a=\frac{-2+\sqrt{4+4\times 9\times 2}}{4}=\frac{-2+\sqrt{76}}{4}\approx \frac{6.7}{4}\approx 1.4$

Hence, $a\in (1,2]$

So, on comparing with $a\in (n-1,n]$ we get $n=2$