Find the angle between the lines $\;$ $\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z - 4}{6}$ $\;$ and $\;$ $x + 1 = \dfrac{y + 2}{2} = \dfrac{z - 4}{2}$
The two lines are
$\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z - 4}{6}$ $\;\;\; \cdots \; (1)$
$\dfrac{x + 1}{1} = \dfrac{y + 2}{2} = \dfrac{z - 4}{2}$ $\;\;\; \cdots \; (2)$
Let $a_1$, $b_1$ and $c_1$ be the direction ratios of line $(1)$.
Let $a_2$, $b_2$ and $c_2$ be the direction ratios of line $(2)$.
Then, $\;$ $a_1 = 2$, $b_1 = 3$, $c_1 = 6$; $\;$ $a_2 = 1$, $b_2 = 2$, $c_2 = 2$
Let $\theta$ be the angle between the lines.
Then,
$\begin{aligned}
\theta & = \cos^{-1} \left[\dfrac{a_1a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right] \\\\
& = \cos^{-1} \left[\dfrac{2 \times 1 + 3 \times 2 + 6 \times 2}{\sqrt{\left(2\right)^2 + \left(3\right)^2 + \left(6\right)^2 } \sqrt{\left(1\right)^2 + \left(2\right)^2 + \left(2\right)^2}}\right] \\\\
& = \cos^{-1} \left[\dfrac{2 + 6 + 12}{\sqrt{4 + 9+ 36} \sqrt{1 + 4 + 4}}\right] \\\\
& = \cos^{-1} \left[\dfrac{20}{\sqrt{49} \sqrt{9}}\right] \\\\
& = \cos^{-1} \left[\dfrac{20}{21}\right]
\end{aligned}$