(i) To find the value of \(p\) for which angle \(AOB\) is \(90^\circ\), we use the dot product formula. The dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\) when the vectors are perpendicular.

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (5)(2) + (1)(7) + (2)(p) = 10 + 7 + 2p = 0\)

Solving for \(p\):

\(17 + 2p = 0\)

\(2p = -17\)

\(p = -\frac{17}{2}\)

(ii) When \(p = 4\), the vector \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2\mathbf{i} + 7\mathbf{j} + 4\mathbf{k}) - (5\mathbf{i} + \mathbf{j} + 2\mathbf{k})\)

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

The modulus of \(\overrightarrow{AB}\) is \(\sqrt{(-3)^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\)

The unit vector in the direction of \(\overrightarrow{AB}\) is \(\frac{-3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}}{7}\)

To find the vector with magnitude 28 in the same direction, multiply the unit vector by 28:

\(28 \times \frac{-3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}}{7} = -12\mathbf{i} + 24\mathbf{j} + 8\mathbf{k}\)