If $ω$ is a complex cube root of unity, then the value of the determinant

$Δ=\begin{bmatrix}1 & ω &ω^2\\ω & ω^2 & 1\\ω^2 & 1 & ω\end{bmatrix},$ is

0

The correct answer is option (2) : 0

Applying $C_1→C_1+C_2+C_3<$ we have

$Δ=\begin{bmatrix}1+ω+ω^2 & ω &ω^2\\ω+ω^2+1 & ω^2 & 1\\ω^2+1+ω & 1 & ω\end{bmatrix}$

$⇒Δ=\begin{bmatrix}0 & ω &ω^2\\0 & ω^2 & 1\\0 & 1 & ω\end{bmatrix}$                 $[∵1+ω+ω^2=0]$

$⇒Δ=0$                 $[∵C_1 $ consists of all zeroes $]$