(i) To find the cosine of angle \(AOB\), we use the dot product formula:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = 3 \times 6 + 0 \times (-3) + (-4) \times 2 = 18 - 8 = 10\).

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{3^2 + 0^2 + (-4)^2} = 5\) and \(|\overrightarrow{OB}| = \sqrt{6^2 + (-3)^2 + 2^2} = 7\).

Thus, \(\cos \angle AOB = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}| |\overrightarrow{OB}|} = \frac{10}{35} = \frac{2}{7}\).

(ii) The vector \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ 6 \end{pmatrix}\).

The magnitude of \(\overrightarrow{AB}\) is \(\sqrt{3^2 + (-3)^2 + 6^2} = \sqrt{9 + 9 + 36} = 7\).

Given \(|\overrightarrow{AB}| = |\overrightarrow{OC}|\), we have:

\(\sqrt{k^2 + (-2k)^2 + (2k - 3)^2} = 7\).

Squaring both sides, \(k^2 + 4k^2 + (2k - 3)^2 = 49\).

\(k^2 + 4k^2 + 4k^2 - 12k + 9 = 49\).

\(9k^2 - 12k - 40 = 0\).

Solving the quadratic equation, \(k = 3\) or \(k = -\frac{5}{3}\).