Answer: (i) The shortest distance between the lines is \(\sqrt6\).
(ii) The acute angle is \(\sin^{-1}\left(\dfrac{\sqrt{42}}{14}\right)\), approximately \(27.6^\circ\) or \(0.481\) radians.
For \(l_1\),
\(\overrightarrow{AB}=(-1+3,5-1,9-4)=(2,4,5)\).
For \(l_2\),
\(\overrightarrow{CD}=(-1+2,7-6,5-5)=(1,1,0)\).
(i) The direction of the common perpendicular is
\((2,4,5)\times(1,1,0)=(-5,5,-2)\).
Using \(\overrightarrow{AC}=(-2+3,6-1,5-4)=(1,5,1)\), the shortest distance is
\(\displaystyle \frac{|(1,5,1)\cdot(-5,5,-2)|}{\sqrt{(-5)^2+5^2+(-2)^2}}=\frac{18}{\sqrt{54}}=\sqrt6\).
(ii) The plane through \(A,B,D\) has direction vectors
\(\overrightarrow{AB}=(2,4,5)\), and \(\overrightarrow{AD}=(-1+3,7-1,5-4)=(2,6,1)\).
A normal vector to the plane is
\((2,4,5)\times(2,6,1)=(-26,8,4)\), so we can use \((-13,4,2)\).
The direction vector of \(l_2\) is \((1,1,0)\). If \(\theta\) is the acute angle between the line and the plane, then
\(\displaystyle \sin\theta=\frac{|(1,1,0)\cdot(-13,4,2)|}{\sqrt{1^2+1^2}\sqrt{(-13)^2+4^2+2^2}}\).
Thus
\(\displaystyle \sin\theta=\frac{9}{\sqrt2\sqrt{189}}=\frac{\sqrt{42}}{14}\).
Therefore
\(\displaystyle \theta=\sin^{-1}\left(\frac{\sqrt{42}}{14}\right)\approx27.6^\circ\).
