$$
\begin{aligned}
& \text { }\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & 2 a+b+c & b \\
c & a & a+2 b+c
\end{array}\right|=2 \\
& \Rightarrow 2(a+b+c)\left|\begin{array}{ccc}
1 & a & b \\
1 & b+c+2 a & b \\
1 & a & c+a+2 b
\end{array}\right|=2
\end{aligned}
$$
[Applying $C_1 \rightarrow C_1+C_2+C_3$ and taking $2(a+b+c)$ common from $\left.C_1\right]$
$$
\Rightarrow 2(a+b+c)\left|\begin{array}{ccc}
1 & a & b \\
0 & b+c+a & 0 \\
0 & 0 & c+a+b
\end{array}\right|=2
$$
[Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$ ]
$$
\begin{aligned}
& \Rightarrow \quad 2(a+b+c)^3=2 \quad \text { [expanding along } C_1 \text { ] } \\
& \Rightarrow \quad(a+b+c)^3=1 \quad \Rightarrow \quad a+b+c=1 \\
&
\end{aligned}
$$
Now, $a^3+b^3+c^3-3 a b c$
$$
\begin{aligned}
& =(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =1 \cdot\left[a^2+b^2+c^2-a b-b c-c a\right] \\
& =(a+b+c)^2-2 a b-2 b c-2 c a-a b-b c-c a \\
& =1-2 a b-2 b c-2 c a-a b-b c-c a \\
& =1-3 a b-3 b c-3 c a
\end{aligned}
$$