​$$\text { Find the angle between the lines } \frac{x-3}{1}=\frac{y-2}{2}=\frac{z+1}{2} \text { and } \frac{x-0}{3}=\frac{y-5}{2}=\frac{z-2}{6} \text {. }$$​​

Correct option is C

​$$\begin{aligned}&\textbf{1. Line 1 } (L_1): \\&\quad \frac{x - 3}{1} = \frac{y - 2}{2} = \frac{z + 1}{2} \\&\quad \text{Direction Vector } (\vec{d}_1): (1, 2, 2) \\[10pt]&\textbf{2. Line 2 } (L_2): \\&\quad \frac{x - 0}{3} = \frac{y - 5}{2} = \frac{z - 2}{6} \\&\quad \text{Direction Vector } (\vec{d}_2): (3, 2, 6) \\[12pt]&\textbf{Step 1: Compute the Dot Product of Direction Vectors} \\&\vec{d}_1 \cdot \vec{d}_2 = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19 \\[10pt]&\textbf{Step 2: Compute the Magnitudes of Direction Vectors} \\&|\vec{d}_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \\&|\vec{d}_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \\[10pt]&\textbf{Step 3: Calculate the Angle Between the Lines} \\&\text{The angle } \theta \text{ between two lines is given by:} \\&\quad \cos \theta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1| \cdot |\vec{d}_2|} \\[5pt]&\text{Substituting the values:} \\&\quad \cos \theta = \frac{19}{3 \times 7} = \frac{19}{21} \\&\quad \theta = \cos^{-1}\left(\frac{19}{21}\right)\end{aligned}$$​​