Answer
The plane with $y=\dfrac{\sqrt 3x}{3}$.
Work Step by Step
Given: $\theta=\dfrac{\pi}{6}$
This angle implies that the surface will be a plane passing through the origin with an angle $\dfrac{\pi}{6}$ in the xy plane and above the x-axis.
Since, $\tan \theta=\dfrac{y}{x}$
and $\tan \dfrac{\pi}{6}=\dfrac{y}{x}$
This implies that $\dfrac{y}{x}=\dfrac{\sqrt 3}{3}$
or, $y=\dfrac{\sqrt 3x}{3}$
Thus, it represents an equation of a plane with $y=\dfrac{\sqrt 3x}{3}$.
