Answer: (i) \(\overrightarrow{AB}=(3-1,\,4-1,\,5-3)=(2,3,2)\). Also \(\overrightarrow{CD}=(0-(-1),\,m-0,\,4-3)=(1,m,1)\). If the lines are parallel, these direction vectors must be proportional. Setting \((1,m,1)=\lambda(2,3,2)\) gives \(\lambda=\tfrac12\) from the first and third components, so \(m=3\lambda=\tfrac32\). Hence the lines are parallel when \(m=\tfrac32\).

(ii) For \(m\neq\tfrac32\), take \(\overrightarrow{AC}=(-1-1,\,0-1,\,3-3)=(-2,-1,0)\). A vector perpendicular to both lines is \(\mathbf n=\overrightarrow{AB}\times\overrightarrow{CD}\): \(\mathbf n=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&3&2\\1&m&1\end{vmatrix}=(3-2m,\,0,\,2m-3)\). The shortest distance between skew lines is \(d=\dfrac{|\overrightarrow{AC}\cdot\mathbf n|}{|\mathbf n|}\). Now \(\overrightarrow{AC}\cdot\mathbf n=(-2)(3-2m)=2(2m-3)\), so \(|\overrightarrow{AC}\cdot\mathbf n|=2|2m-3|\). Also \(|\mathbf n|=\sqrt{(3-2m)^2+(2m-3)^2}=\sqrt{2}\,|2m-3|\). Therefore \(d=\dfrac{2|2m-3|}{\sqrt2|2m-3|}=\sqrt2\).

(iii) When \(m=2\), \(\overrightarrow{CD}=(1,2,1)\). A normal to plane \(ABC\) is \(\overrightarrow{AB}\times\overrightarrow{AC}\): \(\mathbf n_1=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&3&2\\-2&-1&0\end{vmatrix}=(2,-4,4)\sim(1,-2,2)\). A normal to plane \(ABD\) is \(\overrightarrow{AB}\times\overrightarrow{AD}\), where \(\overrightarrow{AD}=(-1,1,1)\): \(\mathbf n_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&3&2\\-1&1&1\end{vmatrix}=(1,-4,5)\). The acute angle between the planes equals the acute angle between their normals, so \(\cos\theta=\dfrac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}=\dfrac{|1+8+10|}{\sqrt{1^2+(-2)^2+2^2}\,\sqrt{1^2+(-4)^2+5^2}}=\dfrac{19}{\sqrt{378}}\). Hence \(\theta\approx 12.2^\circ\).

(i) The direction vector of \(AB\) is

\(\overrightarrow{AB}=(3-1,\,4-1,\,5-3)=(2,3,2)\).

The direction vector of \(CD\) is

\(\overrightarrow{CD}=(0-(-1),\,m-0,\,4-3)=(1,m,1)\).

For the lines to be parallel, these vectors must be scalar multiples. Since the first and third components of \(\overrightarrow{AB}\) are both 2 and those of \(\overrightarrow{CD}\) are both 1, the scale factor is \(\tfrac12\). So

\(m=\tfrac12\cdot 3=\tfrac32\).

Hence \(AB\\parallel CD\) when \(m=\tfrac32\).

(ii) Let \(A=(1,1,3)\) and \(C=(-1,0,3)\). Then

\(\overrightarrow{AC}=(-2,-1,0)\).

A vector perpendicular to both lines is given by the cross product of their direction vectors:

\(\mathbf n=\overrightarrow{AB}\times\overrightarrow{CD}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&3&2\\1&m&1\end{vmatrix}=(3-2m,\,0,\,2m-3)\).

The shortest distance between the skew lines is the component of \(\overrightarrow{AC}\) in the direction of \(\mathbf n\):

\(d=\dfrac{|\overrightarrow{AC}\cdot\mathbf n|}{|\mathbf n|}\).

Now

\(\overrightarrow{AC}\cdot\mathbf n=(-2)(3-2m)+(-1)\cdot 0+0(2m-3)=-2(3-2m)\),

so

\(|\overrightarrow{AC}\cdot\mathbf n|=2|2m-3|\).

Also

\(|\mathbf n|=\sqrt{(3-2m)^2+0^2+(2m-3)^2}=\sqrt{2(2m-3)^2}=\sqrt2\,|2m-3|\).

Therefore

\(d=\dfrac{2|2m-3|}{\sqrt2\,|2m-3|}=\sqrt2\),

which is independent of \(m\) as long as \(m\neq \tfrac32\).

(iii) When \(m=2\), we have \(D=(0,2,4)\), so

\(\overrightarrow{AD}=(-1,1,1)\).

A normal to plane \(ABC\) is

\(\mathbf n_1=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&3&2\\-2&-1&0\end{vmatrix}=(2,-4,4)\),

so we may use \(\mathbf n_1=(1,-2,2)\).

A normal to plane \(ABD\) is

\(\mathbf n_2=\overrightarrow{AB}\times\overrightarrow{AD}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&3&2\\-1&1&1\end{vmatrix}=(1,-4,5)\).

The acute angle between the planes equals the acute angle between these normals. Hence

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

Now

\(\mathbf n_1\cdot\mathbf n_2=1\cdot1+(-2)(-4)+2\cdot5=19\),

\(|\mathbf n_1|=\sqrt{1^2+(-2)^2+2^2}=3,\\quad |\mathbf n_2|=\sqrt{1^2+(-4)^2+5^2}=\sqrt{42}\).

So

\(\cos\theta=\dfrac{19}{3\sqrt{42}}=\dfrac{19}{\sqrt{378}}\),

giving

\(\theta\approx 12.2^\circ\).