(i) To find the points of intersection, substitute \(y = 3 - x\) from the line equation into the curve equation:

\(2x^2 - 8x + 9 = 3 - x\)

Rearrange to form a quadratic equation:

\(2x^2 - 7x + 6 = 0\)

Factorize:

\((2x - 3)(x - 2) = 0\)

Thus, \(x = 2\) or \(x = \frac{3}{2}\).

(ii) To find the stationary point of C, differentiate \(y = 2x^2 - 8x + 9\):

\(\frac{dy}{dx} = 4x - 8\)

Set \(\frac{dy}{dx} = 0\):

\(4x - 8 = 0\)

\(x = 2\)

Therefore, \(x = 2\) is a stationary point of C.