If $a,b,c$ are sides of the triangle $ABC$ and $\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=\mathrm{0,}$ then the value of $\cos 2A+\cos 2B+\cos 2C$ is equal to

By expansion of the determinant, we get,
$\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=1\left(c^2-ab\right)-1\left(ac-b^2\right)+1\left(a^2-bc\right)=0$
$\Rightarrow a^2+b^2+c^2-ab-bc-ca=0$

which is true for $a=b=c$

Hence, $∆ABC$ must be an equilateral triangle
$\Rightarrow A=B=C={60}^{\mathrm{o}}$
$\Rightarrow \cos 2A+\cos 2B+\cos 2C=-\frac{3}{2}$