Answer: (i) The plane is \(x-9y-7z=-4\).

The point of intersection of \(l\) with \(\Pi\) has position vector \(4\mathbf{i}-3\mathbf{j}+5\mathbf{k}\).

(ii) The perpendicular distance from \(P\) to \(\Pi\) is \(\dfrac{21}{\sqrt{131}}\).

(iii) The acute angle between \(l\) and \(\Pi\) is \(23.6^\circ\) (approximately).

Let the two direction vectors of the plane be \(\mathbf{a}=(2,1,-1)\) and \(\mathbf{b}=(1,-3,4)\). A normal to the plane is given by their cross product:

\(\mathbf{n}=\mathbf{a}\times\mathbf{b}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\\\2&1&-1\\\\1&-3&4\end{vmatrix}=\mathbf{i}-9\mathbf{j}-7\mathbf{k}.\)

So the plane has equation \(x-9y-7z=d\) for some constant \(d\). Using the point \((3,0,1)\) on the first line,

\(3-9(0)-7(1)=-4\),

hence \(\Pi: x-9y-7z=-4\).

Now the line \(l\) through \(P(6,-2,1)\) with direction vector \((2,1,-4)\) can be written as

\(x=6+2\lambda,\quad y=-2+\lambda,\quad z=1-4\lambda.\)

To find where it meets the plane, substitute into \(x-9y-7z=-4\):

\((6+2\lambda)-9(-2+\lambda)-7(1-4\lambda)=-4.\)

Simplifying,

\(6+2\lambda+18-9\lambda-7+28\lambda=-4\)

\(17+21\lambda=-4\)

\(21\lambda=-21\)

\(\lambda=-1.\)

So the point of intersection is

\(x=6+2(-1)=4,\quad y=-2+(-1)=-3,\quad z=1-4(-1)=5,\)

therefore the position vector is \(4\mathbf{i}-3\mathbf{j}+5\mathbf{k}\).

For the perpendicular distance from \(P\) to the plane, use

\(\text{distance}=\frac{|x_0-9y_0-7z_0+4|}{\sqrt{1^2+(-9)^2+(-7)^2}}\)

with \(P=(6,-2,1)\):

\(\frac{|6-9(-2)-7(1)+4|}{\sqrt{131}}=\frac{|6+18-7+4|}{\sqrt{131}}=\frac{21}{\sqrt{131}}.\)

So the distance is \(\dfrac{21}{\sqrt{131}}\).

For the angle between the line and the plane, let \(\theta\) be the acute angle between them. Then \(\theta\) is complementary to the angle between the line direction vector \(\mathbf{d}=(2,1,-4)\) and the normal \(\mathbf{n}=(1,-9,-7)\). Hence

\(\mathbf{d}\cdot\mathbf{n}=|\mathbf{d}|\\,|\mathbf{n}|\sin\theta.\)

Now

\(\mathbf{d}\cdot\mathbf{n}=2(1)+1(-9)+(-4)(-7)=21,\)

and

\(|\mathbf{d}|=\sqrt{2^2+1^2+(-4)^2}=\sqrt{21},\quad |\mathbf{n}|=\sqrt{1^2+(-9)^2+(-7)^2}=\sqrt{131}.\)

So

\(21=\sqrt{21}\\,\sqrt{131}\\,\sin\theta,\)

giving

\(\sin\theta=\sqrt{\frac{21}{131}}.\)

Therefore \(\theta\approx 23.6^\circ\).