Given $\left|\begin{array}{lll}\mathrm{b} & 1 & \mathrm{a} \\ \mathrm{a} & 1 & \mathrm{c} \\ \mathrm{c} & 1 & \mathrm{~b}\end{array}\right|=0$ b expand w.r.t ' $\mathrm{R}_1$,
$\begin{aligned} & \Rightarrow \mathrm{b}(\mathrm{b}-\mathrm{c})-1\left(\mathrm{ba}-\mathrm{c}^2\right)+\mathrm{a}(\mathrm{a}-\mathrm{c})=0 \\ & \mathrm{~b}^2-\mathrm{bc}-\mathrm{ba}+\mathrm{c}^2+\mathrm{a}^2-\mathrm{ac}=0\end{aligned}$


Here, $\mathrm{ABC}$ is the triangle with sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$.
From (i), Multiply and Divide by 2 ,
$$
\begin{aligned}
& \Rightarrow \frac{1}{2}\left[2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a\right]=0 \\
& \Rightarrow \frac{1}{2}\left[\left(a^2-2 a b+b^2\right)+\left(a^2+c^2-2 c a\right)+\left(b^2+c^2-2 b c\right)\right]=0 \\
& \Rightarrow \frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0
\end{aligned}
$$
So, $\mathrm{a}=\mathrm{b}=\mathrm{c}$, then $\triangle \mathrm{ABC}$ is an equilateral triangle.
So, $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60^{\circ}$
Then, $2(\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C})=2 \times 3 \times \cos 60^{\circ}$
$$
=6 \times \frac{1}{2}=3
$$
So, option (c) is correct.