$\omega$ is root of $x+\frac{1}{x}+1=0$, i.e. $x^2+x+1=0$
This gives roots $\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm i \sqrt{3}}{2}$
$\therefore \omega=\frac{-1+\sqrt{-3}}{2}$ and $\omega^2=\frac{-1-\sqrt{-3}}{2}$ are roots of $x^2+x+1=0$
Also, $\omega^3=\omega \cdot \omega^2=1$ and $1+\omega+\omega^2=0$
Then, $A=\left|\begin{array}{ccc}1 & 1+\omega & 1+\omega+\omega^2 \\ 3 & 4+3 \omega & 5+4 \omega+3 \omega^2 \\ 6 & 9+6 \omega & 11+9 \omega+6 \omega^2\end{array}\right|$ is written as,

$$
\begin{aligned}
& |A|=\left|\begin{array}{ccc}
1 & 1+\omega & 1+\omega+\omega^2 \\
3 & 4+3 \omega & 3\left(1+\omega+\omega^2\right)+2+\omega \\
6 & 9+6 \omega & 6\left(1+\omega+\omega^2\right)+5+3 \omega
\end{array}\right| \\
& |A|=\left|\begin{array}{ccc}
1 & 1+\omega & 0 \\
3 & 4+3 \omega & 2+\omega \\
6 & 9+6 \omega & 5+3 \omega
\end{array}\right|
\end{aligned}
$$

Expand along row 1,
$$
\begin{aligned}
& |A|=(4+3 \omega)(5+3 \omega)-(2+\omega)(9+6 \omega) \\
& -(1+\omega)\{15+9 \omega-12-6 \omega\} \\
& =20+27 \omega+9 \omega^2-18-21 \omega-6 \omega^2 \\
& -3-6 \omega-3 \omega^2=-1
\end{aligned}
$$