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

Question

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

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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$$⁡θ$$+$$$$cos$$⁡$$2$$θ$$+$$$$cos$$⁡$$3$$θ$$=0$$ and $$sin$$⁡θ−$$sin$$⁡$$2$$θ$$+$$$$sin$$⁡$$3$$θ$$=0or$$  or $$cos$$⁡$$2$$θ$$(2cos$$⁡θ$$+1)=0$$ and $$sin$$⁡$$2$$θ$$(1$$−$$2cos$$⁡θ$$)=0which$$ is not possible as $$cos$$⁡$$2$$θ$$=0$$ gives $$sin$$⁡$$2$$θ≠$$0$$,$$cos$$⁡θ≠$$1/2$$. And $$cos$$⁡θ$$=$$−$$1/2$$ gives $$sin$$⁡$$2$$θ≠$$0$$,$$cos$$⁡θ≠$$1/2$$. . Therefore, Eq. $$(1) does not hold simultaneously. Therefore,$$a+b+c$$≠$$0$$⇒$$a=b=c$$∴eiθ$$=e$$−$$2i$$θ$$=e3i$$θwhich is satisfied only by eiθ$$=1$$, i.e., $$cos$$⁡θ$$=1$$,$$sin$$⁡θ$$=0$$ so θ$$=2k$$π,k∈Z.

If $a = \cos θ + isin θ, b = \cos 2 θ - isin 2 θ, c = \cos 3 θ$ $+ isin 3 θ and if |\begin{array}{lll} a b c \\ b c a \\ c a b \end{array}| = 0$ , then

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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

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If $a = \cos θ + isin θ, b = \cos 2 θ - isin 2 θ, c = \cos 3 θ$ $+ isin 3 θ and if |\begin{array}{lll} a b c \\ b c a \\ c a b \end{array}| = 0$ , then