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.

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.

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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

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$θ = (4 k + 1) π, k \in Z$

none of these

$θ = 2 kπ, k \in Z$

$θ = (2 k + 1) π, k \in Z$

answer is A.

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Detailed Solution

$\begin{array}{l} Δ = |\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}| \\ = - (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca) \\ = - \frac{1}{2} (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0 \\ \Rightarrow a + b + c = 0 or a = b = c \end{array}$

$If a + b + c = 0, we have$

$\cos θ + \cos 2 θ + \cos 3 θ = 0 and \sin θ - \sin 2 θ + \sin 3 θ = 0$

or $or \cos 2 θ (2 \cos θ + 1) = 0 and \sin 2 θ (1 - 2 \cos θ) = 0$

which is not possible as $\cos 2 θ = 0$ gives $\sin 2 θ \neq 0, \cos θ \neq 1 / 2$ . And $\cos θ = - 1 / 2$ gives $\sin 2 θ \neq 0, \cos θ \neq 1 / 2 .$ . Therefore, Eq. (1) does not hold simultaneously. Therefore,