(i) To verify that the angles \(DAB\), \(DAC\), and \(CAB\) are \(90^\circ\), we check the dot products:

\(\overrightarrow{AB} \cdot \overrightarrow{AC} = 3 \times 1 + 1 \times (-2) + 1 \times (-1) = 3 - 2 - 1 = 0\)

\(\overrightarrow{AB} \cdot \overrightarrow{AD} = 3 \times 1 + 1 \times 4 + 1 \times (-7) = 3 + 4 - 7 = 0\)

\(\overrightarrow{AC} \cdot \overrightarrow{AD} = 1 \times 1 + (-2) \times 4 + (-1) \times (-7) = 1 - 8 + 7 = 0\)

Since all dot products are zero, the angles are \(90^\circ\).

(ii) The area of triangle \(ABC\) is given by:

\(\text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right|\)

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

\(= \mathbf{i}(1 + 2) - \mathbf{j}(-3 - 1) + \mathbf{k}(-6 - 1)\)

\(= 3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}\)

\(\left| \overrightarrow{AB} \times \overrightarrow{AC} \right| = \sqrt{3^2 + 4^2 + (-7)^2} = \sqrt{66}\)

\(\text{Area} = \frac{1}{2} \sqrt{66}\)

The volume of the pyramid is:

\(V = \frac{1}{3} \times \frac{1}{2} \sqrt{66} \times \sqrt{66} = \frac{1}{6} \times 66 = 11\)

(i) The vector \(\overrightarrow{XP} = -4\mathbf{i} + (p-5)\mathbf{j} + 2\mathbf{k}\). For \(\angle OPX = 90^0\), the dot product \(\overrightarrow{OP} \cdot \overrightarrow{XP} = 0\). This gives:

\([-4\mathbf{i} + (p-5)\mathbf{j} + 2\mathbf{k}] \cdot (pj + 2k) = 0\)

\(p^2 - 5p + 4 = 0\)

Solving the quadratic equation, \(p = 1 \text{ or } 4\).

(ii) For \(p = 9\), \(\overrightarrow{XP} = -4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\). The magnitude \(|\overrightarrow{XP}| = \sqrt{16 + 16 + 4} = 6\).

The unit vector is \(\frac{1}{6}(-4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})\).

(iii) \(\overrightarrow{AG} = -4\mathbf{i} + 15\mathbf{j} + 2\mathbf{k}\). Since \(\overrightarrow{XQ} = \lambda \overrightarrow{AG}\), and \(\lambda = \frac{2}{3}\),

\(\overrightarrow{XQ} = -\frac{8}{3}\mathbf{i} + \frac{10}{3}\mathbf{j} + \frac{4}{3}\mathbf{k}\).

(i) Since \(CP = \frac{3}{5} CA\), and \(CA = 4\mathbf{i} - 3\mathbf{k}\), we have:

\(\overrightarrow{CP} = \frac{3}{5}(4\mathbf{i} - 3\mathbf{k}) = 2.4\mathbf{i} - 1.8\mathbf{k}\).

(ii) \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}\). Since \(AP = \frac{3}{5} \times 4\mathbf{i} = 2.4\mathbf{i}\),

\(\overrightarrow{OP} = 2.4\mathbf{i} + 1.2\mathbf{k}\).

\(\overrightarrow{BP} = \overrightarrow{OP} - \overrightarrow{OB} = (2.4\mathbf{i} + 1.2\mathbf{k}) - 2.4\mathbf{j} = 2.4\mathbf{i} - 2.4\mathbf{j} + 1.2\mathbf{k}\).

(iii) Use the scalar product formula:

\(\overrightarrow{BP} \cdot \overrightarrow{CP} = 2.4 \times 2.4 + (-2.4) \times 0 + 1.2 \times (-1.8) = 5.76 - 2.16 = 3.6\).

\(|\overrightarrow{BP}| = \sqrt{2.4^2 + (-2.4)^2 + 1.2^2} = \sqrt{12.96}\),

\(|\overrightarrow{CP}| = \sqrt{2.4^2 + (-1.8)^2} = \sqrt{9}\).

\(\cos BPC = \frac{3.6}{\sqrt{12.96} \times \sqrt{9}} = \frac{1}{3}\).

Angle BPC = \(\cos^{-1}(\frac{1}{3}) \approx 70.5^{\circ}\) (or 1.23 radians).

(i) To find the unit vector in the direction of \(\overrightarrow{PM}\), we first calculate \(\overrightarrow{PM}\) as follows:

\(\overrightarrow{PM} = 2\mathbf{i} - 10\mathbf{k} + \frac{1}{2}(6\mathbf{j} + 8\mathbf{k})\)

Simplifying, \(\overrightarrow{PM} = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\).

The magnitude of \(\overrightarrow{PM}\) is \(\sqrt{4 + 9 + 36} = 7\).

Thus, the unit vector is \(\frac{1}{7} (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})\).

(ii) For \(\angle ATP = \cos^{-1}\left(\frac{2}{7}\right)\), we use the vectors:

\(\overrightarrow{AT} = 6\mathbf{j} + 8\mathbf{k}\)

\(\overrightarrow{PT} = a\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\)

\(\cos(ATP) = \frac{(6\mathbf{j} + 8\mathbf{k}) \cdot (a\mathbf{i} + 6\mathbf{j} - 2\mathbf{k})}{\sqrt{36 + 64}\sqrt{a^2 + 36 + 4}}\)

\(= \frac{36 - 16}{\sqrt{100}\sqrt{a^2 + 40}} = \frac{20}{10\sqrt{a^2 + 40}}\)

\(\frac{2}{\sqrt{a^2 + 40}} = \frac{2}{7}\)

Solving, \(a = 3\).

(i) To find \(\overrightarrow{AM}\), note that \(M\) is the midpoint of \(OX\). Since \(OX = 10\mathbf{k}\), \(M\) is at \(5\mathbf{k}\). \(A\) is at \(8\mathbf{i}\). Thus, \(\overrightarrow{AM} = 5\mathbf{k} - 8\mathbf{i} = -6\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}\).

For \(\overrightarrow{AC}\), \(C\) is at \(8\mathbf{j}\). Thus, \(\overrightarrow{AC} = 8\mathbf{j} - 8\mathbf{i} = -8\mathbf{i} + 8\mathbf{j}\).

(ii) The scalar product \(\overrightarrow{AM} \cdot \overrightarrow{AC} = (-6)(-8) + (2)(8) + (5)(0) = 48 + 16 = 64\).

The magnitudes are \(|\overrightarrow{AM}| = \sqrt{(-6)^2 + 2^2 + 5^2} = \sqrt{65}\) and \(|\overrightarrow{AC}| = \sqrt{(-8)^2 + 8^2} = \sqrt{128}\).

Using the cosine formula, \(64 = \sqrt{128} \sqrt{65} \cos \theta\).

Solving for \(\cos \theta\), we find \(\theta = 45.4^\circ\).