Equation of line passing through origin and making $30^\circ, 60^\circ$ and $90^\circ$ with $x, y, z$ axes, respectively is

$\frac{2x}{\sqrt{3}} = \frac{2y}{1} = \frac{z}{0}$

The correct answer is Option (2) → $\frac{2x}{\sqrt{3}} = \frac{2y}{1} = \frac{z}{0}$ ##

$\cos \alpha = \cos 30^\circ = \frac{\sqrt{3}}{2}$

$\cos \beta = \cos 60^\circ = \frac{1}{2}$

$\cos \gamma = \cos 90^\circ = 0$

Equation of required line

$\frac{x-0}{\frac{\sqrt{3}}{2}} = \frac{y-0}{\frac{1}{2}} = \frac{z-0}{0}$

$\frac{2x}{\sqrt{3}} = \frac{2y}{1} = \frac{z}{0}$