$${L_1}$$ and $${L_2}$$ are two lines whose vector equations are
$$\eqalign{ & {L_1}:\overrightarrow r = \lambda \left( {\left( {\cos \,\theta + \sqrt 3 } \right)\hat i + \left( {\sqrt 2 \,\sin \,\theta } \right)\hat j + \left( {\cos \,\theta - \sqrt 3 } \right)\hat k} \right) \cr & {L_2}:\overrightarrow r = \mu \left( {a\hat i + b\hat j + c\hat k} \right), \cr} $$
where $$\lambda $$ and $$\mu $$ are scalars and $$\alpha $$ is the acute angle between $${L_1}$$ and $${L_2}.$$ If the angle $$'\alpha '$$ is independent of $$\theta $$ then the value of $$'\alpha '$$ is :

Solution :
Both the lines pass through origin. Line $${L_1}$$ is parallel to the vector
$$\overrightarrow {{V_1}} = \left( {\cos \,\theta + \sqrt 3 } \right)\hat i + \left( {\sqrt 2 \,\sin \,\theta } \right)\hat j + \left( {\cos \,\theta - \sqrt 3 } \right)\hat k$$
and $${L_2}$$ is parallel to the vector
$$\eqalign{ & \overrightarrow {{V_2}} = a\hat i + b\hat j + c\hat k \cr & \therefore \,\cos \,\alpha = \frac{{\overrightarrow {{V_1}} .\overrightarrow {{V_2}} }}{{\left| {\overrightarrow {{V_1}} } \right|\left| {\overrightarrow {{V_2}} } \right|}} \cr & = \frac{{a\left( {\cos \,\theta + \sqrt 3 } \right) + \left( {b\sqrt 2 } \right)\sin \,\theta + c\left( {\cos \,\theta - \sqrt 3 } \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{\left( {\cos \,\theta + \sqrt 3 } \right)}^2} + 2\,{{\sin }^2}\theta + {{\left( {\cos \,\theta - \sqrt 3 } \right)}^2}} }} \cr & = \frac{{\left( {a + c} \right)\cos \,\theta + \left( {b\sqrt 3 } \right) + \sin \,\theta + \left( {a - c - \sqrt 3 } \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {2 + 6} }} \cr} $$
In order that $$\cos \,\alpha $$   is independent of $$\theta ,$$  we get $$a + c = 0$$   and $$b = 0$$
$$\therefore \,\cos \,\alpha = \frac{{2a\sqrt 3 }}{{a\sqrt 2 \,2\sqrt 2 }} = \frac{{\sqrt 3 }}{2}\, \Rightarrow \alpha = \frac{\pi }{6}$$