Explanation

(a)
Both the lines pass through origin. Line $L_{1}$ is parallel to the vector ${\overrightarrow{V}}_{1}$
${\overrightarrow{V}}_{1} = \left( \cos\theta + \sqrt{3} \right)\widehat{i} + \left( \sqrt{2}\sin\theta \right)\widehat{j} + (\cos\theta - \sqrt{3})\widehat{k}$
and $L_{2}$ is parallel to the vector ${\overrightarrow{V}}_{2}$
${\overrightarrow{V}}_{2} = a\widehat{i} + b\widehat{j} + c\widehat{k}$
$\therefore\ \cos\alpha = \frac{{\overrightarrow{V}}_{1}.{\overrightarrow{V}}_{2}}{\left| {\overrightarrow{V}}_{1} \right||{\overrightarrow{V}}_{2}|}$
$= \frac{a\left( \cos\theta + \sqrt{3} \right) + (b\sqrt{2})\sin\theta + c(\cos\theta - \sqrt{3})}{\sqrt{a^{2} + b^{2} + c^{2}}\ \sqrt{\left( \cos\theta + \sqrt{3} \right)^{2} + 2\sin^{2}\theta + \left( \cos\theta - \sqrt{3} \right)^{2}}}$
$= \frac{(a + c)\cos\theta + b\sqrt{2}\sin\theta + (a - c)\sqrt{3}}{\sqrt{a^{2} + b^{2} + c^{2}}\sqrt{2 + 6}}$
For $\cos\alpha$ to be independent of $\theta$, we get
$a + c = 0$ and $b = 0$
$\therefore\ \cos\alpha = \frac{2a\sqrt{3}}{a\sqrt{2}\ 2\sqrt{2}} = \frac{\sqrt{3}}{2}$
$\Rightarrow \ \alpha = \frac{\pi}{6}$