If a=cos⁡θ+isin⁡θ,b=cos⁡2θ−isin⁡2θ,c=cos⁡3θ +isin⁡3θ and if a    b    cb    c    ac    a    b=0, then

Δ=abcbcacab  =−(a+b+c)a2+b2+c2−ab−bc−ca  =−12(a+b+c)(a−b)2+(b−c)2+(c−a)2=0⇒a+b+c=0 or a=b=c If a+b+c=0, we have cos⁡θ+cos⁡2θ+cos⁡3θ=0 and sin⁡θ−sin⁡2θ+sin⁡3θ=0or  or cos⁡2θ(2cos⁡θ+1)=0 and sin⁡2θ(1−2cos⁡θ)=0which is not possible as cos⁡2θ=0 gives sin⁡2θ≠0,cos⁡θ≠1/2. And cos⁡θ=−1/2 gives sin⁡2θ≠0,cos⁡θ≠1/2. . Therefore, Eq. (1) does not hold simultaneously. Therefore,a+b+c≠0⇒a=b=c∴eiθ=e−2iθ=e3iθwhich is satisfied only by eiθ=1, i.e., cos⁡θ=1,sin⁡θ=0 so θ=2kπ,k∈Z.