The angle between the line $2x = 3y = z$ and x-axis is:

$\cos^{-1}(\frac{3}{7})$

The correct answer is Option (2) → $\cos^{-1}(\frac{3}{7})$

Given line: $\frac{2x}{1} = \frac{3y}{1} = \frac{z}{1}$

Rewriting in symmetric form: $\frac{x}{\frac{1}{2}} = \frac{y}{\frac{1}{3}} = \frac{z}{1}$

So, the direction ratios of the line are:

$\left( \frac{1}{2}, \frac{1}{3}, 1 \right)$

The direction vector of the x-axis is $\vec{i} = (1, 0, 0)$

Let $\vec{a} = \left( \frac{1}{2}, \frac{1}{3}, 1 \right)$ and $\vec{b} = (1, 0, 0)$

The angle $\theta$ between vectors $\vec{a}$ and $\vec{b}$ is given by:

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

$\vec{a} \cdot \vec{b} = \frac{1}{2}(1) + \frac{1}{3}(0) + 1(0) = \frac{1}{2}$

$|\vec{a}| = \sqrt{\left( \frac{1}{2} \right)^2 + \left( \frac{1}{3} \right)^2 + 1^2} = \sqrt{ \frac{1}{4} + \frac{1}{9} + 1 } = \sqrt{ \frac{49}{36} } = \frac{7}{6}$

$|\vec{b}| = 1$

$\cos \theta = \frac{\frac{1}{2}}{\frac{7}{6}} = \frac{3}{7}$

$\theta = \cos^{-1} \left( \frac{3}{7} \right)$