Answer: (a) The shortest distance between the lines is \(\frac{16}{3}\).

(b) The acute angle between \(\Pi_1\) and \(\Pi_2\) is \(36.7^\circ\).

(a) First find direction vectors for the two lines.

For \(l_1\), \(\overrightarrow{AB}=(8-2,\,7-3,\,-13-(-5))=(6,4,-8)\), so a direction vector is \((3,2,-4)\).

For \(l_2\), \(\overrightarrow{CD}=(3-(-2),\,-1-1,\,4-8)=(5,-2,-4)\).

A vector perpendicular to both lines is given by the cross product

\(\;(3,2,-4)\times(5,-2,-4)=(-16,-8,-16)\),

so we may use the simpler parallel vector \((2,1,2)\) as the common perpendicular direction.

Now take any vector joining a point on \(l_1\) to a point on \(l_2\), for example \(\overrightarrow{AD}\):

\(\overrightarrow{AD}=(3-2,\,-1-3,\,4-(-5))=(1,-4,9)\).

The shortest distance between the skew lines is the component of \(\overrightarrow{AD}\) in the common perpendicular direction:

\(\displaystyle d=\frac{|\overrightarrow{AD}\cdot(2,1,2)|}{\sqrt{2^2+1^2+2^2}}=\frac{|2-4+18|}{3}=\frac{16}{3}.\)

(b) To find the angle between the planes, find a normal to each plane.

For \(\Pi_1\), use \(\overrightarrow{AD}=(1,-4,9)\) and \(\overrightarrow{AB}=(3,2,-4)\):

\(\overrightarrow{n_1}=\overrightarrow{AD}\times\overrightarrow{AB}=(-2,31,14).\)

For \(\Pi_2\), use \(\overrightarrow{AD}=(1,-4,9)\) and \(\overrightarrow{CD}=(5,-2,-4)\):

\(\overrightarrow{n_2}=\overrightarrow{AD}\times\overrightarrow{CD}=(34,49,18).\)

The acute angle between the planes is the angle between these normals, so

\(\displaystyle \cos\theta=\frac{\overrightarrow{n_1}\cdot\overrightarrow{n_2}}{|\overrightarrow{n_1}|\,|\overrightarrow{n_2}|}= \frac{(-2)(34)+31\cdot49+14\cdot18}{\sqrt{(-2)^2+31^2+14^2}\,\sqrt{34^2+49^2+18^2}}.\)

Evaluating gives \(\theta\approx 36.7^\circ\).