Correct Answer - 1
Let `A=[(1" "omega" "omega^2),(omega" "omega^2" "1),(omega^2" " 1 " "z+omega)]`
`A=[(0" "0" "0),(0" "0" "0),(0" "0" "0)]`and Tr (A)=0,|A|=0
`A^3=0`
`A=[(z+1" "omega" "omega^3),(omega" "z+omega^2" "1),(omega^2" "1" "z+omega)]`=[A+zl]=0
`rArr " "z^3=0`
`rArr z=0 ` the number of z satisfying the given equation is 1.