(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:

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