The measure of the angle between the lines \(x = k + 1, y = 2k - 1, z = 2k + 3, k \in \mathbb{R}\) and \(\dfrac{x - 1}{2} = \dfrac{y - 2}{1} = \dfrac{z - 3}{1}\) is

The equation of the first line is \(x = k + 1, y = 2k - 1, z = 2k + 3\).

This can be written in symmetric form as \(\dfrac{x - 1}{1} = \dfrac{y + 1}{2} = \dfrac{z - 3}{2} = k\).

The direction ratios of the first line are \(1, 2, 2\).

The equation of the second line is \(\dfrac{x - 1}{2} = \dfrac{y - 2}{1} = \dfrac{z - 3}{1}\).

The direction ratios of the second line are \(2, 1, 1\).

Let \(\theta\) be the angle between the two lines. Then,

\(\cos \theta = \dfrac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}\)

\(\cos \theta = \dfrac{|(1)(2) + (2)(1) + (2)(1)|}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + 1^2 + 1^2}}\)

\(\cos \theta = \dfrac{6}{\sqrt{9} \sqrt{6}} = \dfrac{6}{3\sqrt{6}} = \dfrac{2}{\sqrt{6}} = \sqrt{\dfrac{2}{3}}\)

\(\theta = \cos^{-1}\left(\sqrt{\dfrac{2}{3}}\right)\)

Answer: \(\cos^{-1}\left(\sqrt{\dfrac{2}{3}}\right)\)