(i) Differentiate the function: \(\frac{dy}{dx} = 3x^2 - 6x - 9\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(3x^2 - 6x - 9 = 0\).

Factorize: \(3(x^2 - 2x - 3) = 0\).

Solve: \(x^2 - 2x - 3 = 0\) gives \((x - 3)(x + 1) = 0\).

Thus, \(x = 3\) or \(x = -1\).

Since the minimum point is on the \(x\)-axis, \(y = 0\) at \(x = 3\).

Substitute \(x = 3\) into the original equation: \(3^3 - 3(3)^2 - 9(3) + k = 0\).

Simplify: \(27 - 27 - 27 + k = 0\).

Thus, \(k = 27\).

(ii) Substitute \(x = -1\) into the original equation to find the maximum point:

\(y = (-1)^3 - 3(-1)^2 - 9(-1) + 27\).

Simplify: \(y = -1 - 3 + 9 + 27 = 32\).

Thus, the maximum point is \((-1, 32)\).

(iii) The function is decreasing where \(\frac{dy}{dx} < 0\).

From \(3x^2 - 6x - 9 = 0\), the roots are \(x = 3\) and \(x = -1\).

Test intervals: \(-1 < x < 3\) is where \(\frac{dy}{dx} < 0\).

Thus, the function is decreasing for \(-1 < x < 3\).