(i) The equation of the line with gradient \(-2\) passing through \(P(3t, 2t)\) is:

\(y - 2t = -2(x - 3t)\)

Rearranging gives:

\(y = -2x + 6t + 2t\)

\(y = -2x + 8t\)

To find \(A\), set \(y = 0\):

\(0 = -2x + 8t\)

\(x = 4t\)

So, \(A(4t, 0)\).

To find \(B\), set \(x = 0\):

\(y = 8t\)

So, \(B(0, 8t)\).

The area of triangle \(AOB\) is:

\(\text{Area} = \frac{1}{2} \times 4t \times 8t = 16t^2\)

(ii) The line through \(P\) perpendicular to \(AB\) has gradient \(\frac{1}{2}\). Its equation is:

\(y - 2t = \frac{1}{2}(x - 3t)\)

Rearranging gives:

\(y = \frac{1}{2}x - \frac{3}{2}t + 2t\)

\(y = \frac{1}{2}x + \frac{1}{2}t\)

To find \(C\), set \(y = 0\):

\(0 = \frac{1}{2}x + \frac{1}{2}t\)

\(x = -t\)

So, \(C(-t, 0)\).

The midpoint of \(PC\) is:

\(\left( \frac{3t - t}{2}, \frac{2t + 0}{2} \right) = (t, t)\)

This lies on the line \(y = x\).