${\mathrm{L}}_1\text{ and }{\mathrm{L}}_2$ are two lines whose vector equations are ${\mathrm{L}}_1 : \overline{\mathrm{r}}=\mathrm{\lambda }\left(\right(\cos \mathrm{\theta }+\sqrt{3})\overline{\mathrm{i}}+(\sqrt{2}\sin \mathrm{\theta })\overline{\mathrm{j}}+(\cos \mathrm{\theta }-\sqrt{3}\left)\overline{\mathrm{k}}\right)$ ; ${\mathrm{L}}_2 : \overline{\mathrm{r}}=\mathrm{\mu }(\mathrm{a}\overline{\mathrm{i}}+\mathrm{b}\overline{\mathrm{j}}+\mathrm{c}\overline{\mathrm{k}})$

Where $\mathrm{\lambda }\text{ and }\mathrm{\mu }$ are scalars and $\mathrm{\alpha }$ is the acute angle between ${\mathrm{L}}_1\text{ and }{\mathrm{L}}_2$. If the angle $\mathrm{\alpha }$ (is independent of $\mathrm{\theta }$ ) is equal to $\frac{\mathrm{\pi }}{\mathrm{k}}$ then k = _____

Both the $\text{ lines passes through the origin line }{\mathrm{L}}_1\text{ is }$$\text{ parallel to the vector }$

${\overline{\mathrm{V}}}_1=(\cos \mathrm{\theta }+\sqrt{3})\overline{\mathrm{i}}+(\sqrt{2}\sin \mathrm{\theta })\overline{\mathrm{j}}+(\cos \theta -\sqrt{3})\overline{\mathrm{k}}$

And ${\mathrm{L}}_2$ is parallel to the vector ${\overline{\mathrm{V}}}_2=\mathrm{a}\overline{\mathrm{i}}+\mathrm{b}\overline{\mathrm{j}}+\mathrm{c}\overline{\mathrm{k}}$

$\begin{array}{l}\cos \mathrm{\alpha }=\frac{\mathrm{a}(\cos \mathrm{\theta }+\sqrt{3})+\mathrm{b}(\sqrt{2}\sin \mathrm{\theta })+\mathrm{c}(\cos \mathrm{\theta }-\sqrt{3}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}\sqrt{(\cos \mathrm{\theta }+\sqrt{3}{)}^2+(\sqrt{2\sin \mathrm{\theta }}{)}^2(\cos \mathrm{\theta }-\sqrt{3}{)}^2}}\\ =\frac{(\mathrm{a}+\mathrm{c})\cos \mathrm{\theta }+\mathrm{b}\sqrt{2}\sin \mathrm{\theta }+(\mathrm{a}-\mathrm{c})\sqrt{3}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}\cdot \sqrt{8}}\end{array}$

$\therefore \mathrm{\alpha }\text{ is independent of }\mathrm{\theta }$

$\begin{array}{l}\Rightarrow \mathrm{a}+\mathrm{c}=0.\mathrm{b}=0\\ \cos \mathrm{\alpha }=\frac{(\mathrm{a}-\mathrm{c})\sqrt{3}}{\sqrt{2}\mathrm{a}\cdot 2\sqrt{2}}=\frac{2\mathrm{a}\sqrt{3}}{\sqrt{2}\mathrm{a}2\sqrt{2}}=\frac{\sqrt{3}}{2};\mathrm{\alpha }=\frac{\mathrm{\pi }}{6}\end{array}$