(i) To show that \(AB\) is perpendicular to \(BC\), we calculate the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}\)

Calculate the dot product:

\(\overrightarrow{AB} \cdot \overrightarrow{BC} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} = (-2)(3) + (-1)(2) + (2)(4) = -6 - 2 + 8 = 0\)

Since the dot product is zero, \(AB\) is perpendicular to \(BC\).

(ii) To find the position vector of \(D\), first find the unit vector in the direction of \(CD\):

\(\overrightarrow{CD} = \pm 4 \times \overrightarrow{BA} = \pm 4 \times \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \\ -8 \end{pmatrix}\)

Given \(CD = 12\), the position vector \(\overrightarrow{OD}\) is:

\(\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} + \begin{pmatrix} 8 \\ 4 \\ -8 \end{pmatrix} = \begin{pmatrix} 12 \\ 9 \\ -2 \end{pmatrix}\)

(a) To find \(\overrightarrow{MD}\), we first find \(\overrightarrow{OM}\) which is the midpoint of \(AB\). Since \(OA = 4\mathbf{i}\) and \(OB = 4\mathbf{j}\), \(OM = 2\mathbf{i} + 2\mathbf{j}\). Thus, \(\overrightarrow{MD} = \overrightarrow{OD} - \overrightarrow{OM} = 3\mathbf{k} - (2\mathbf{i} + 2\mathbf{j}) = -2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\).

For \(\overrightarrow{ON}\), since \(N\) divides \(BC\) in the ratio 2:1, \(\overrightarrow{ON} = \overrightarrow{OB} + \frac{2}{3}(\overrightarrow{OC} - \overrightarrow{OB}) = 4\mathbf{j} + \frac{2}{3}(\mathbf{i} + \mathbf{k} - 4\mathbf{j}) = 3\mathbf{j} + \mathbf{k}\).

(b) The angle \(\theta\) between \(\overrightarrow{MD}\) and \(\overrightarrow{ON}\) is given by \(\cos \theta = \frac{\overrightarrow{MD} \cdot \overrightarrow{ON}}{|\overrightarrow{MD}| |\overrightarrow{ON}|}\).

\(\overrightarrow{MD} \cdot \overrightarrow{ON} = (-2)(0) + (-2)(3) + (3)(1) = -6 + 3 = -3\).

\(|\overrightarrow{MD}| = \sqrt{(-2)^2 + (-2)^2 + 3^2} = \sqrt{17}\), \(|\overrightarrow{ON}| = \sqrt{3^2 + 1^2} = \sqrt{10}\).

\(\cos \theta = \frac{-3}{\sqrt{17} \cdot \sqrt{10}}\).

\(\theta = \cos^{-1}\left(\frac{-3}{\sqrt{170}}\right) \approx 103.3^\circ\).

(c) To find the perpendicular distance from \(M\) to \(ON\), consider the projection of \(\overrightarrow{OM}\) on \(\overrightarrow{ON}\).

\(\text{Projection of } \overrightarrow{OM} \text{ on } \overrightarrow{ON} = \frac{\overrightarrow{OM} \cdot \overrightarrow{ON}}{|\overrightarrow{ON}|^2} \overrightarrow{ON}\).

\(\overrightarrow{OM} \cdot \overrightarrow{ON} = (2\mathbf{i} + 2\mathbf{j}) \cdot (3\mathbf{j} + \mathbf{k}) = 6\).

\(\text{Projection length} = \frac{6}{10} = \frac{3}{5}\).

\(\text{Perpendicular distance} = \sqrt{|\overrightarrow{OM}|^2 - \left(\frac{3}{5}\right)^2 |\overrightarrow{ON}|^2} = \sqrt{8 - \frac{9}{5}} = \sqrt{\frac{22}{5}}\).

(i) Since D is the midpoint of AB, the coordinates of D are \((4, 3, 0)\) because A is at \((8, 0, 0)\) and B is at \((0, 6, 0)\). Thus, \(\overrightarrow{OD} = 4\mathbf{i} + 3\mathbf{j}\).

For \(\overrightarrow{CD}\), since C is at \((0, 0, 10)\), \(\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = 4\mathbf{i} + 3\mathbf{j} - 10\mathbf{k}\).

(ii) To find \(\angle ODC\), use the scalar product formula:

\(\overrightarrow{OD} \cdot \overrightarrow{CD} = |\overrightarrow{OD}| \times |\overrightarrow{CD}| \times \cos \theta\)

\(\overrightarrow{OD} \cdot \overrightarrow{CD} = (4)(4) + (3)(3) + (0)(-10) = 16 + 9 = 25\)

The magnitudes are \(|\overrightarrow{OD}| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\) and \(|\overrightarrow{CD}| = \sqrt{4^2 + 3^2 + (-10)^2} = \sqrt{125}\).

Thus, \(25 = 5 \times \sqrt{125} \times \cos \theta\).

Solving for \(\cos \theta\), we get \(\cos \theta = \frac{25}{5 \times \sqrt{125}} = \frac{1}{\sqrt{5}}\).

Therefore, \(\theta = \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \approx 63.4^\circ\) or 1.11 radians.

(i) To find \(\overrightarrow{DB}\), note that \(\overrightarrow{OB} = 6\mathbf{i} + 4\mathbf{j}\) and \(\overrightarrow{OD} = 3\mathbf{k}\). Therefore, \(\overrightarrow{DB} = \overrightarrow{OB} - \overrightarrow{OD} = (6\mathbf{i} + 4\mathbf{j}) - 3\mathbf{k} = 6\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}\).

For \(\overrightarrow{DE}\), since \(E\) is the center of the rectangle, \(\overrightarrow{OE} = 3\mathbf{i} + 2\mathbf{j}\). Thus, \(\overrightarrow{DE} = \overrightarrow{OE} - \overrightarrow{OD} = (3\mathbf{i} + 2\mathbf{j}) - 3\mathbf{k} = 3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\).

(ii) To find angle \(\theta\) between \(\overrightarrow{DB}\) and \(\overrightarrow{DE}\), use the dot product: \(\overrightarrow{DB} \cdot \overrightarrow{DE} = 6 \times 3 + 4 \times 2 + (-3) \times (-3) = 18 + 8 + 9 = 35\).

The magnitudes are \(|\overrightarrow{DB}| = \sqrt{6^2 + 4^2 + (-3)^2} = \sqrt{61}\) and \(|\overrightarrow{DE}| = \sqrt{3^2 + 2^2 + (-3)^2} = \sqrt{22}\).

Using the formula \(\overrightarrow{DB} \cdot \overrightarrow{DE} = |\overrightarrow{DB}| |\overrightarrow{DE}| \cos \theta\), we have:

\(35 = \sqrt{61} \times \sqrt{22} \times \cos \theta\).

Solving for \(\cos \theta\), we get \(\cos \theta = \frac{35}{\sqrt{61} \times \sqrt{22}}\).

Finally, \(\theta = \cos^{-1}\left(\frac{35}{\sqrt{61} \times \sqrt{22}}\right) \approx 17.2^\circ\) (0.300 rad).

(i) First, find \(\overrightarrow{OC}\) using \(\overrightarrow{OC} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).

Calculate the dot product \(\overrightarrow{OC} \cdot \overrightarrow{OB} = 2 \times 5 + (-3) \times 0 + 6 \times 2 = 22\).

Find the magnitudes: \(|\overrightarrow{OC}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7\) and \(|\overrightarrow{OB}| = \sqrt{5^2 + 0^2 + 2^2} = \sqrt{29}\).

Use the cosine formula: \(\overrightarrow{OC} \cdot \overrightarrow{OB} = |\overrightarrow{OC}| \cdot |\overrightarrow{OB}| \cdot \cos \angle BOC\).

\(22 = 7 \times \sqrt{29} \times \cos \angle BOC\).

\(\cos \angle BOC = \frac{22}{7 \times \sqrt{29}}\).

\(\angle BOC = \cos^{-1}\left(\frac{22}{7 \times \sqrt{29}}\right) \approx 54.3^\circ\) (or 0.948 rad).

(ii) The magnitude of \(\overrightarrow{OC}\) is 7. To find a vector with magnitude 35, scale \(\overrightarrow{OC}\) by \(\frac{35}{7} = 5\).

Thus, the vector is \(\pm 5 \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).