$\because A^{T}A=I$

$\Rightarrow \left[\begin{array}{lll}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]\left[\begin{array}{lll}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{lll}{a}^2+b^2+c^2& ab+bc+ac& ab+bc+ac\\ ab+bc+ac& a^2+b^2+c^2& ab+bc+ac\\ ab+bc+ac& ab+bc+ac& a^2+b^2+c^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

On comparing each element both sides, we get

$a^2+b^2+c^2=1 \& ab+bc+ca=0 .........\left(\mathrm{i}\right)$

We know that $a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)$.

$\Rightarrow 2-3abc=\left(a+b+c\right)\left(1-0\right)$ (from equation $\left(\mathrm{i}\right)$)

$\Rightarrow 2-3abc=\left(a+b+c\right) .........\left(\mathrm{ii}\right)$

Now, we also know that ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)$.

$\Rightarrow {\left(a+b+c\right)}^2=1+2\left(0\right)$ (from equation $\left(\mathrm{i}\right)$)

$\Rightarrow a+b+c=\pm 1$

Putting it in equation $\left(\mathrm{ii}\right)$, we get

$2-3abc=\pm 1$

$\Rightarrow 3abc=2\mp 1$

$\Rightarrow abc=\frac{1}{3} \text{or} abc=1$