Let  $a,b,c\in R$ be all non-zero and satisfy $a^3+b^3+c^3=2$. If the matrix  $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$ satisfies  $A^{T}A=I$, the value of abc can be:

$A^{T}A=I\Rightarrow \left[\begin{array}{l}\sum a^2\sum ab\sum ab\\ \sum ab\sum a^2\sum ab\\ \sum ab\sum ab\sum a^2\end{array}\right]=\left[\begin{array}{l}100\\ 010\\ 001\end{array}\right]$

$\Rightarrow \sum a^2=1;\sum ab=0$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=a+b+c=\sqrt{{(a+b+c)}^2}=\sqrt{\sum a^2-2\sum ab}$

$=\pm 1$

$2-3abc=1\Rightarrow abc=1/3$

$or2-3abc=-1\Rightarrow abc=1$