If $A=\begin{bmatrix}1&2&-1\\-1&1&2\\2&-1&1\end{bmatrix}$, then $det (adj (adj\, A))$, is

$14^4$

We know that for a square matrix of order n $adj (adj\, A) =|A|^{n-2} A$, if $|A|≠0$.

$⇒det (adj (adj\, A)) = ||A|^{n-2} A|$

$⇒det (adj (adj\, A)) =(|A|^{n-2})^n|A|$

$⇒det (adj (adj\, A)) =|A|^{n^2 - 2n +1}$

Here, $n = 3$ and $|A|= 14$

$∴ det (adj (adj\, A)) = 14^4$