The angle between the lines $l_1:\frac{x+1}{1}=\frac{2-y}{2}=\frac{z-1}{1}$ and $l_2:\frac{x-1}{4}=\frac{2y-4}{6}=\frac{z-1}{2}$ is

$\frac{\pi}{2}$

The correct answer is Option (4) → $\frac{\pi}{2}$

$l_{1} : \frac{x+1}{1} = \frac{2-y}{2} = \frac{z-1}{1}$

$\text{Direction ratios of } l_{1}: (1,\,-2,\,1)$

$l_{2} : \frac{x-1}{4} = \frac{2y-4}{6} = \frac{z-1}{2}$

$\text{Direction ratios of } l_{2}: (4,\,3,\,2)$

$\cos\theta = \frac{1\cdot4 + (-2)\cdot3 + 1\cdot2}{\sqrt{1^{2}+(-2)^{2}+1^{2}}\;\sqrt{4^{2}+3^{2}+2^{2}}}$

$\cos\theta = \frac{4 - 6 + 2}{\sqrt{6}\,\sqrt{29}}$

$\cos\theta = 0$

$\theta = \frac{\pi}{2}$