The acute angle between the lines $\vec r = (4\hat i −\hat j) +λ(2\hat i +\hat j-3\hat k)$ and $\frac{x − 1}{1} = \frac{y + 1}{-3} = \frac{z − 2}{2}$ is

$\frac{\pi}{3}$

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

Line 1 direction vector:

$\vec{d}_{1} = \langle 2,\,1,\,-3\rangle$

Line 2 in symmetric form:

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

Direction vector:

$\vec{d}_{2} = \langle 1,\,-3,\,2\rangle$

Use formula:

$\cos\theta = \frac{\vec{d}_{1}\cdot \vec{d}_{2}}{|\vec{d}_{1}|\;|\vec{d}_{2}|}$

Dot product:

$\vec{d}_{1}\cdot \vec{d}_{2} = 2\cdot1 + 1\cdot(-3) + (-3)\cdot2$

$= 2 - 3 - 6 = -7$

Magnitudes:

$|\vec{d}_{1}| = \sqrt{2^{2}+1^{2}+(-3)^{2}} = \sqrt{14}$

$|\vec{d}_{2}| = \sqrt{1^{2}+(-3)^{2}+2^{2}} = \sqrt{14}$

Thus:

$\cos\theta = \frac{-7}{14} = -\frac{1}{2}$

Acute angle means take the acute angle between the lines:

$\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$

The acute angle between the two lines is $\frac{\pi}{3}$.