The angle between two lines whose direction ratios are proportional to 1, 1, -2 and $(\sqrt{3}-1),(-\sqrt{3}-1),-4$ is:

$\frac{\pi}{3}$

The correct answer is Option (1) →  $\frac{\pi}{3}$

let $\vec{v_1}=\hat i+\hat j-2\hat k$

$\vec{v_2}=(\sqrt{3}-1)\hat i+(-\sqrt{3}-1)\hat j-4\hat k$

$|\vec{v_1}|=\sqrt{1^2+1^2+(-2)^2}=\sqrt{6}$

$|\vec{v_2}|=\sqrt{(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2+(-4)^2}$

$=\sqrt{3+1-2\sqrt{3}+3+1+2\sqrt{3}+16}$

$=\sqrt{24}=2\sqrt{6}$

so $\vec{v_1},\vec{v_2}=|\vec{v_1}||\vec{v_2}|\cos θ$

$\cos θ=\frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6}×2\sqrt{6}}$

$=\frac{6}{2×6}=\frac{1}{2}$

$θ=\frac{\pi}{3}$