(i) To find \(\overrightarrow{CD}\), calculate \(3\mathbf{a} - 2\mathbf{b}\):

\(3\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} - 2\begin{pmatrix} 4 \\ 0 \\ 6 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} - \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).

The magnitude of \(\overrightarrow{CD}\) is \(\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7\).

Thus, the unit vector is \(\frac{1}{7} \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).

(ii) The position vector of E is the midpoint of C and D:

\(\overrightarrow{OE} = \frac{1}{2}(3\mathbf{a} + 2\mathbf{b}) = \frac{1}{2}\left(\begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} + \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix}\right) = \begin{pmatrix} 7 \\ 1.5 \\ 9 \end{pmatrix}\).

Calculate \(\overrightarrow{OE} \cdot \overrightarrow{OD}\):

\(\begin{pmatrix} 7 \\ 1.5 \\ 9 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix} = 56 + 0 + 108 = 164\).

The magnitudes are \(|\overrightarrow{OE}| = \sqrt{7^2 + 1.5^2 + 9^2} = \sqrt{132.25} \approx 11.5\) and \(|\overrightarrow{OD}| = \sqrt{8^2 + 0^2 + 12^2} = \sqrt{208}\).

Using the dot product formula, \(164 = 11.5 \times \sqrt{208} \times \cos \theta\).

Solving for \(\theta\), \(\theta = \cos^{-1}\left(\frac{164}{11.5 \times \sqrt{208}}\right) \approx 8.6^\circ\).