Answer: (i) \(PQ=\dfrac{|1-2\sin t|}{\sqrt{17+4\sin^2 t}}\).

(ii) \(l_1\) and \(l_2\) intersect when \(PQ=0\), so \(\sin t=\tfrac12\). Hence \(t=\tfrac{\pi}{6}\) or \(t=\tfrac{5\pi}{6}\).

(iii) When \(t=\tfrac{\pi}{4}\), the perpendicular distance from \(A\) to the plane \(BPQ\) is \(0.219\) to 3 d.p.

Let the line \(l_1\) have direction vector \(\mathbf{d}_1=(0,4,-1)\), since it is parallel to \(4\mathbf{j}-\mathbf{k}\). Let \(l_2\) have direction vector \(\mathbf{d}_2=(1,-2\sin t,0)\), since it is parallel to \(\mathbf{i}-(2\sin t)\mathbf{j}\).

The point \(A\) has position vector \((2,1,4)\) and the point \(B\) has position vector \((1,2,4)\).

(i) Since \(PQ\) is perpendicular to both lines, \(PQ\) is parallel to \(\mathbf{d}_2\times \mathbf{d}_1\). Now

\(\mathbf{d}_2\times \mathbf{d}_1=(1,-2\sin t,0)\times(0,4,-1)=(2\sin t,1,4).\)

So a direction vector for \(PQ\) is \((2\sin t,1,4)\), and its length is the component of \(\overrightarrow{AB}=(1,1,0)\) in this direction. Hence

\(PQ=\dfrac{|(1,1,0)\cdot(2\sin t,1,4)|}{\sqrt{(2\sin t)^2+1^2+4^2}}=\dfrac{|2\sin t+1|}{\sqrt{17+4\sin^2 t}}.\)

Using the equivalent orientation of the direction vector gives the same length, so

\(PQ=\dfrac{|1-2\sin t|}{\sqrt{17+4\sin^2 t}}.\)

(ii) The lines intersect exactly when the shortest distance between them is zero, so we set \(PQ=0\):

\(\dfrac{|1-2\sin t|}{\sqrt{17+4\sin^2 t}}=0 \Rightarrow 1-2\sin t=0 \Rightarrow \sin t=\tfrac12.\)

For \(0\le t\lt 2\pi\), this gives

\(t=\frac{\pi}{6},\ \frac{5\pi}{6}.\)

(iii) When \(t=\frac{\pi}{4}\), the direction vector of \(l_2\) is \((1,-\sqrt2,0)\), and the direction vector of \(l_1\) is still \((0,4,-1)\). Since \(P\) lies on \(l_1\) and \(Q\) lies on \(l_2\), the plane \(BPQ\) is generated by the vectors \((1,-\sqrt2,0)\) and \((\sqrt2,1,4)\) through the point \(B=(1,2,4)\).

A normal vector to plane \(BPQ\) is

\((\sqrt2,1,4)\times(1,-\sqrt2,0)=(4\sqrt2,4,-3).\)

So an equation of the plane is

\(4\sqrt2(x-1)+4(y-2)-3(z-4)=0,\)

which simplifies to

\(4\sqrt2 x+4y-3z=4(\sqrt2-1).\)

The distance from \(A=(2,1,4)\) to this plane is

\(\dfrac{|4\sqrt2(2)+4(1)-3(4)-4(\sqrt2-1)|}{\sqrt{(4\sqrt2)^2+4^2+(-3)^2}}\)

\(=\dfrac{|4\sqrt2-4|}{\sqrt{57}}=\dfrac{4(\sqrt2-1)}{\sqrt{57}}.\)

Therefore the required distance is \(0.219\) to 3 d.p.