The angle at which the line, $\frac{x-1}{0}=\frac{2-y}{-1}=\frac{2z-3}{-2}$ is inclined with the positive direction of z-axis is

$\frac{3\pi}{4}$

The correct answer is Option (4) → $\frac{3\pi}{4}$

Given line: $\frac{x-1}{0}=\frac{2-y}{-1}=\frac{2z-3}{-2}$

Direction ratios: $(0,-1,-2)$

Angle $\theta$ with positive z-axis is given by:

$\cos\theta = \frac{\text{z-component}}{\text{magnitude of direction vector}}$

$\Rightarrow \cos\theta = \frac{-2}{\sqrt{0^2+(-1)^2+(-2)^2}}=\frac{-2}{\sqrt{5}}$

$\cos\theta$ is negative, so $\theta$ is an obtuse angle.

$\theta = \pi - \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)$

Approximation gives $\theta=\frac{3\pi}{4}$

$\theta=\frac{3\pi}{4}$