Answer: (i) \(PQ=\dfrac{4}{\sqrt{3}}\)

(ii) \(\Pi:\ 4x+5y+z=-15\)

(iii) The perpendicular distance of \(A\) from \(\Pi\) is \(\dfrac{38}{\sqrt{42}}\)

Let the direction vectors of the lines be \(\mathbf d_1=(1,-2,-3)\) and \(\mathbf d_2=(-2,1,3)\).

(i) Length of \(PQ\)

Since \(PQ\) is perpendicular to both lines, it is parallel to \(\mathbf d_1\times\mathbf d_2\):

\(\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&-2&-3\\-2&1&3\end{vmatrix}=(-3,3,-3)\),

so a parallel vector is \((1,-1,1)\).

Also, with \(A=(3,3,-4)\) and \(B=(-3,-1,2)\),

\(\overrightarrow{BA}=A-B=(6,4,-6).\)

The shortest distance between the skew lines is the projection of \(\overrightarrow{BA}\) onto this perpendicular direction:

\(PQ=\left|\frac{(6,4,-6)\cdot(1,-1,1)}{\sqrt{1^2+(-1)^2+1^2}}\right|=\left|\frac{6-4-6}{\sqrt3}\right|=\frac{4}{\sqrt3}.\)

(ii) Cartesian equation of the plane \(\Pi\)

The plane contains \(l_2\), so it contains the direction vector \((-2,1,3)\). It also contains \(PQ\), with direction \((1,-1,1)\). Therefore a normal vector to the plane is

\((1,-1,1)\times(-2,1,3)=(-4,-5,-1),\)

so we may take \(\mathbf n=(4,5,1)\).

Since \(B=(-3,-1,2)\) lies on the plane,

\(4x+5y+z=c\), and substituting \((-3,-1,2)\) gives

\(c=4(-3)+5(-1)+2=-15.\)

Hence the plane is

\(\Pi:\ 4x+5y+z=-15.\)

(iii) Perpendicular distance of \(A\) from \(\Pi\)

Using the plane equation \(4x+5y+z+15=0\), the distance from \(A=(3,3,-4)\) is

\(\frac{|4(3)+5(3)+(-4)+15|}{\sqrt{4^2+5^2+1^2}}=\frac{38}{\sqrt{42}}.\)