​$$\text { The angle between the lines } \frac{x-4}{2}=\frac{y}{1}=\frac{z+1}{-2}, \frac{x-1}{4}=\frac{y+1}{-4}=\frac{z-2}{2} \text { is: }$$​​

Correct option is D

​$$\begin{aligned}&\textbf{Given Lines:} \\[5pt]&\text{1. Line 1 (}L_1\text{):} \quad \frac{x - 4}{2} = \frac{y}{1} = \frac{z + 1}{-2} \\&\quad \text{Direction Vector } (\vec{d}_1): (2, 1, -2) \\[8pt]&\text{2. Line 2 (}L_2\text{):} \quad \frac{x - 1}{4} = \frac{y + 1}{-4} = \frac{z - 2}{2} \\&\quad \text{Direction Vector } (\vec{d}_2): (4, -4, 2) \\[12pt]&\textbf{Compute the Dot Product of Direction Vectors} \\&\vec{d}_1 \cdot \vec{d}_2 = (2)(4) + (1)(-4) + (-2)(2) = 8 - 4 - 4 = 0 \\[10pt]&\textbf{Compute the Magnitudes of Direction Vectors} \\&|\vec{d}_1| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \\&|\vec{d}_2| = \sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \\[10pt]&\textbf{Calculate the Angle Between the Lines} \\&\text{The angle } \theta \text{ between two lines is given by:} \\&\cos \theta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1| \cdot |\vec{d}_2|} \\[6pt]&\text{Substituting the values:} \\&\cos \theta = \frac{0}{3 \times 6} = 0 \\&\theta = \cos^{-1}(0) = 90^\circ\end{aligned}$$​​