Đáp án:

Đường thẳng \( AB \) qua \( A(5,1,3) \) có VTCP:  \( \vec{AB} = (-4, 5, -1). \)

Đường thẳng \( CD \) qua \( C(5,0,4) \) có VTCP:  \( \vec{CD} = (-1, 0, 2). \)

\( d(AB, CD) = \frac{|[\vec{AB}, \vec{CD}] \cdot \vec{AC}|}{|[\vec{AB}, \vec{CD}]|}. \)

\( \begin{aligned} \left[\vec{AB}, \vec{CD}\right] = (10, 9, 5) \\ \vec{AC} = (0, -1, 1) \end{aligned} \Rightarrow [\vec{AB}, \vec{CD}] \cdot \vec{AC} = -9 + 5 = -4. \)

\( \Rightarrow d(AB, CD) = \frac{4}{\sqrt{100 + 81 + 25}} = \frac{4}{\sqrt{206}}. \)