Let $A=\left[\begin{array}{ccc}0& 1& 2\\ a& 0& 3\\ 1& c& 0\end{array}\right],$ where $a,c\in R.$ If $A^3=A$ and the positive value of $a$ belongs to the interval $(n-1,n),$ where $n\in N,$ then $n$ is equal to ________

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

$A^3=A$

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

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

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

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

Given $A^3=A$

$2ac+3=\mathrm{0...}\left(1\right)anda+2+3c=1$

$a+1+3c=0$

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

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

$f\left(1\right)<0,f\left(2\right)>0$

$a\in (1,2)$