 Solution

The three columns of $N=\sqrt{8}A$ are $(\sqrt{3},1,2)$, $(-\sqrt{2},\sqrt{6},0)$, and $(-\sqrt{3},-1,2)$. Each has squared Euclidean norm eight. Their pairwise dot products areHence $N^{\mathsf{T}}N=8I$, so $\boxed{A^{\mathsf{T}}A=I}$. Thus $A$ is an orthogonal matrix, with $A^{-1}=A^{\mathsf{T}}$. The determinant from (i) places it in the determinant-one component.