Let ω=−12+32i Then the value of the determinant Δ=1111−1−ω2ω21ω2ω4 is

Using 1+ω2=−ω,ω4=ω and applying  C2→C2−C1,C3→C3−C1 we get Δ=1001ω−1ω2−11ω2−1ω−1=(ω−1)2−ω2−12=ω+ω2−2ω−1−ω2+1=(−3)ω−ω2=3ω(ω−1)