If in a $\Delta ABC$,  ${\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C=3\sin A\sin B\sin C$, then the value of the determinant  $\left|\begin{array}{lll}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ is 

$\left|\begin{array}{lll}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=\left|\begin{array}{lll}a+b+c& a+b+c& a+b+c\\ b& c& a\\ c& a& b\end{array}\right|R_1:R_1+R_2+R_3$

$=(a+b+c)\left|\begin{array}{lll}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|$

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

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

$=-(a^3+b^3+c^3-3abc)$

$=-8R^3({\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C-3\sin A\sin B\sin C)$

$=0$