The values of $α$ for which the vector $\bar{a} = \hat{i} + 3 \hat{j} + (\sin 2 α) \hat{k}$ makes an obtuse angle with the z-axis and the vectors
$\bar{b} = (\tan α) i - j + 2 \sqrt{\sin \frac{α}{2}} \hat{k} and \bar{c} = (\tan α) \hat{i} + (\tan α) \hat{j} - 3 \sqrt{cosec \frac{α}{2}} \hat{k}$ are mutually orthogonal are

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$nπ - \tan^{- 1} 2, n \in I$

$(4 n + 2) π - \tan^{- 1} 2, n \in I$

$2 nπ - \tan^{- 1} 2, n \in I$

$(2 n + 1) π - \tan^{- 1} 2, n \in I$

answer is D.

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Detailed Solution

$\begin{array}{l} \vec{a} \cdot \vec{k} < 0 \Rightarrow \sin 2 α < 0 \\ \vec{b} \cdot \vec{c = 0} \Rightarrow \tan^{2} α - \tan α - 6 = 0 \Rightarrow \tan α = - 2 or 3 \\ \sin 2 α < 0 if \tan α = - 2 \end{array}$
for which $\sin \frac{α}{2} > 0 if α$ ,lies in the second quadrant
$∴ α = π + \tan^{- 1} (- 2) = π - \tan^{- 1} 2$
General value is $α = 2 nπ + π - \tan^{- 1} 2, n \in I$