Cosine of the acute angle between the lines $\frac{x-3}{2}=\frac{y-2}{1}=\frac{z-5}{2}$ and $\frac{x-1}{6}=\frac{y-3}{-3}=\frac{z+6}{2}$ is

$\frac{13}{21}$

The correct answer is Option (2) → $\frac{13}{21}$

The cosine of the angle between two lines is given by the formula:

$\cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$

For line 1: direction ratios = (2, 1, 2)

For line 2: direction ratios = (6, -3, 2)

Numerator: $2 \cdot 6 + 1 \cdot (-3) + 2 \cdot 2 = 12 - 3 + 4 = 13$

Denominator: $\sqrt{2^2 + 1^2 + 2^2} \cdot \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{9} \cdot \sqrt{49} = 3 \cdot 7 = 21$

So, $\cos\theta = \frac{13}{21}$