Given,

 $\left[\begin{array}{ccc}14& 28& -14\\ -14& 14& 28\\ 28& -14& 14\end{array}\right]=\left(adj\left(adjA\right)\right)$

Now we know that $\left|adjA\right|=A^2$ if it is $3\times 3$ matrix

So $\left|adj\left(adjA\right)\right|={\left|adjA\right|}^2$

$\Rightarrow {\left|A\right|}^{2\times 2}={\left|A\right|}^4$

Now, ${\left|A\right|}^4=\left|\begin{array}{ccc}14& 28& -14\\ -14& 14& 28\\ 28& -14& 14\end{array}\right|$

$\Rightarrow {\left|A\right|}^4={14}^3 \left|\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right|$

$\Rightarrow {\left|A\right|}^4={14}^3\times \left(14\right)$

${\left|A\right|}^4={\left(14\right)}^4 \Rightarrow \left|A\right|=14$