Answer: (i) \(\lambda^2+\lambda-20=0\), so \(\lambda=4\) or \(\lambda=-5\).

(ii) The acute angle between the two planes is \(66.1^\circ\) to 3 significant figures.

The direction vector of \(AB\) is

\(\overrightarrow{AB}=\mathbf b-\mathbf a=\begin{pmatrix}4\\3-\lambda\\1\end{pmatrix}.\)

A direction vector of \(CD\) is proportional to

\(\overrightarrow{CD}=\mathbf d-\mathbf c=\begin{pmatrix}0\\5\\5\end{pmatrix}\sim\begin{pmatrix}0\\1\\1\end{pmatrix}.\)

A vector perpendicular to both lines is therefore

\(\mathbf n=\begin{pmatrix}0\\1\\1\end{pmatrix}\times\begin{pmatrix}4\\3-\lambda\\1\end{pmatrix}=\begin{pmatrix}\lambda-2\\4\\-4\end{pmatrix}.\)

Also

\(\overrightarrow{DB}=\mathbf b-\mathbf d=\begin{pmatrix}5\\-4\\-6\end{pmatrix}.\)

The shortest distance between the skew lines is

\(\frac{|\overrightarrow{DB}\cdot\mathbf n|}{|\mathbf n|}=3.\)

So

\(\frac{|5(\lambda-2)-16+24|}{\sqrt{(\lambda-2)^2+16+16}}=3,\)

which gives

\(\frac{|5\lambda-2|}{\sqrt{\lambda^2-4\lambda+36}}=3.\)

Squaring,

\((5\lambda-2)^2=9(\lambda^2-4\lambda+36).\)

Expanding and simplifying:

\(25\lambda^2-20\lambda+4=9\lambda^2-36\lambda+324,\)

so

\(16\lambda^2+16\lambda-320=0.\)

Dividing by \(16\) gives

\(\lambda^2+\lambda-20=0.\)

Hence \(\lambda=4\) or \(\lambda=-5\).

For each value of \(\lambda\), find a normal to the plane through \(A\), \(B\), and \(D\) using \(\overrightarrow{AB}\times\overrightarrow{AD}\).

When \(\lambda=4\), a normal is proportional to

\(\mathbf n_1=\begin{pmatrix}-10\\-29\\11\end{pmatrix}.\)

When \(\lambda=-5\), a normal is proportional to

\(\mathbf n_2=\begin{pmatrix}44\\-29\\56\end{pmatrix}.\)

The angle between the planes is the acute angle between their normals:

\(\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}=\frac{1017}{\sqrt{1062}\sqrt{5913}}.\)

Therefore

\(\theta=66.1^\circ\)

to 3 significant figures.