(a) To find the value of p such that AC and BC are equal in length, we use the distance formula:

\((5 - 2p)^2 + (p + 2)^2 = (10 - 2p)^2 + (3 - p)^2\)

Expanding both sides:

\(25 - 20p + 4p^2 + p^2 + 4p + 4 = 100 - 40p + 4p^2 + 9 - 6p + p^2\)

Simplifying gives:

\(30p = 80\)

Thus, \(p = \frac{8}{3}\).

(b)(i) Given that AC is perpendicular to BC, the slopes \(m_{AC}\) and \(m_{BC}\) must satisfy \(m_{AC} \times m_{BC} = -1\).

\(m_{AC} = \frac{p + 2}{2p - 5}\) and \(m_{BC} = \frac{p - 3}{2p - 10}\)

Setting \(\frac{p + 2}{2p - 5} \times \frac{p - 3}{2p - 10} = -1\) and solving gives:

\(p^2 - p - 6 = 0\)

Factoring gives \((p - 4)(p - 11) = 0\), so \(p = 4\) (ignoring \(p = \frac{11}{5}\) as p is an integer).

(b)(ii) The circle passes through A, B, and C. The midpoint of AB is \((7.5, 0.5)\).

The radius squared \(r^2 = \frac{50}{4}\) or \(r = \frac{5\sqrt{2}}{2}\).

The equation of the circle is \((x - 7.5)^2 + (y - 0.5)^2 = \frac{50}{4}\).

Expanding and simplifying gives:

\(x^2 + y^2 - 15x - y + 44 = 0\).