Answer: \(\Pi_1:\ 2x+3y+6z=14.\)

The values of \(\lambda\) for which the distance from \(P\) to \(\Pi_1\) is not greater than \(4\) are

\(-1\leq\lambda\leq0.\)

The acute angle between \(\Pi_1\) and \(\Pi_2\) is

\(\cos^{-1}\left(\frac1{\sqrt{21}}\right)\approx77.4^\circ.\)

The plane \(\Pi_1\) contains two direction vectors

\((0,2,-1)\quad\text{and}\quad(3,2,-2).\)

A normal vector is

\((0,2,-1)\times(3,2,-2)=(-2,-3,-6).\)

Using the opposite normal \((2,3,6)\), and the point \((1,2,1)\),

\(2(x-1)+3(y-2)+6(z-1)=0.\)

Therefore

\(\Pi_1:\quad 2x+3y+6z=14.\)

A point \(P\) on the line in the question has coordinates

\(P=(3+4\lambda,\ 8+6\lambda,\ 2+5\lambda).\)

The distance from \(P\) to \(\Pi_1\) is

\(\frac{|2(3+4\lambda)+3(8+6\lambda)+6(2+5\lambda)-14|}{\sqrt{2^2+3^2+6^2}}.\)

The denominator is \(7\), and the numerator simplifies to

\(|28+56\lambda|.\)

So the distance is

\(|4+8\lambda|.\)

We require

\(|4+8\lambda|\leq4.\)

This gives

\(-4\leq4+8\lambda\leq4,\)

so

\(-1\leq\lambda\leq0.\)

Now find the angle between \(\Pi_1\) and \(\Pi_2\). A normal to \(\Pi_1\) is

\(\mathbf n_1=(2,3,6).\)

For \(\Pi_2\), use two direction vectors in the plane, for example

\((4,6,5)\quad\text{and}\quad(2,6,1).\)

Their cross product is proportional to

\(\mathbf n_2=(4,-1,-2).\)

The acute angle \(\theta\) between the planes is the acute angle between their normals:

\(\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}.\)

Now

\(\mathbf n_1\cdot\mathbf n_2=2(4)+3(-1)+6(-2)=-7,\)

\(|\mathbf n_1|=7,\qquad |\mathbf n_2|=\sqrt{4^2+(-1)^2+(-2)^2}=\sqrt{21}.\)

Hence

\(\cos\theta=\frac7{7\sqrt{21}}=\frac1{\sqrt{21}}.\)

Therefore

\(\theta=\cos^{-1}\left(\frac1{\sqrt{21}}\right)\approx77.4^\circ.\)