Answer:

The plane \(\Pi_1\) can be written as

\(\mathbf{r}=(4,-2,2)+\lambda(1,7,1)+\mu(3,1,-1).\)

When \(\lambda=0\) and \(\mu=0\), this gives the point \((4,-2,2)\), which is the position vector of \(A\). Hence \(A\) lies in \(\Pi_1\).

A normal vector to \(\Pi_1\), and therefore to the parallel plane \(\Pi_2\), is the cross product of the two direction vectors:

\((1,7,1)\times(3,1,-1)=(-8,4,-20).\)

So a perpendicular vector is any non-zero multiple of this, for example

\((2,-1,5),\)

or \(2\mathbf{i}-\mathbf{j}+5\mathbf{k}\).

Since \(\overrightarrow{OA}=(4,-2,2)\),

\(\overrightarrow{OL}=3\overrightarrow{OA}=(12,-6,6).\)

The plane \(\Pi_2\) has normal vector \((2,-1,5)\), so its equation is

\(2x-y+5z=d.\)

Substituting \(L(12,-6,6)\) gives

\(d=2(12)-(-6)+5(6)=60,\)

so \(\Pi_2\) is \(2x-y+5z=60\).

The point \(N\) lies on the line through \(O\) in the normal direction, so let

\((x,y,z)=t(2,-1,5).\)

Substitute into \(2x-y+5z=60\):

\(2(2t)-(-t)+5(5t)=60,\)

so \(30t=60\), giving \(t=2\). Hence

\(N=(4,-2,10).\)

Therefore the position vector of \(N\) is \(4\mathbf{i}-2\mathbf{j}+10\mathbf{k}\).

Now \(\overrightarrow{AM}=3\overrightarrow{ML}\), so \(AM:ML=3:1\). Thus

\(\overrightarrow{OM}=\overrightarrow{OA}+\frac{3}{4}(\overrightarrow{OL}-\overrightarrow{OA}).\)

Since \(\overrightarrow{OL}-\overrightarrow{OA}=(8,-4,4)\),

\(\overrightarrow{OM}=(4,-2,2)+\frac{3}{4}(8,-4,4)=(10,-5,5).\)

This proves the stated position vector of \(M\).

The line through \(O\) and \(N\) has direction vector \((2,-1,5)\). Also

\(\overrightarrow{NM}=(10,-5,5)-(4,-2,10)=(6,-3,-5).\)

The perpendicular distance from \(M\) to this line is

\(\frac{|\overrightarrow{NM}\times(2,-1,5)|}{|(2,-1,5)|}.\)

Now

\((6,-3,-5)\times(2,-1,5)=(-20,-40,0),\)

so

\(\text{distance}=\frac{20\sqrt5}{\sqrt{30}}=\frac{20}{\sqrt6}\approx8.16.\)