(a) To find p, use the condition that line AB is perpendicular to line BC. The gradient of AB is \(\frac{7 - 4}{p - 6}\) and the gradient of BC is \(\frac{18 - 7}{14 - p}\). Since they are perpendicular, their product is -1:

\(\frac{7 - 4}{p - 6} \times \frac{18 - 7}{14 - p} = -1\)

Solving this equation gives \(p^2 - 20p + 51 = 0\). Solving the quadratic equation, we find \(p = 3\) or \(p = 17\). Given \(p < 10\), the value of \(p\) is 3.

(b) The center of the circle is the midpoint of AC, which is \((10, 11)\). The radius is the distance from the center to any of the points, say A: \(\sqrt{(10 - 6)^2 + (11 - 4)^2} = \sqrt{65}\). Thus, the equation of the circle is \((x - 10)^2 + (y - 11)^2 = 65\).

(c) The gradient of the radius at C is \(\frac{11 - 18}{10 - 14} = \frac{7}{4}\). The tangent at C is perpendicular to this radius, so its gradient is \(-\frac{4}{7}\). Using point-slope form at C (14, 18), the equation is:

\(y - 18 = -\frac{4}{7}(x - 14)\)

Rearranging gives \(4x + 7y - 182 = 0\).