(i) Differentiate the curve: \(\frac{dy}{dx} = 3x^2 - 18x + 24\).

Since the tangent is parallel to the line \(y = 2 - 3x\), the slope of the tangent is \(-3\).

Set \(3x^2 - 18x + 24 = -3\).

Solve for \(x\):

\(3x^2 - 18x + 24 + 3 = 0\)

\(3x^2 - 18x + 27 = 0\)

\(x^2 - 6x + 9 = 0\)

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

\(x = 3\)

Substitute \(x = 3\) into the original equation to find \(y\):

\(y = 3^3 - 9(3)^2 + 24(3) - 12 = 27 - 81 + 72 - 12 = 6\)

The point \(A\) is \((3, 6)\).

The equation of the tangent is \(y - 6 = -3(x - 3)\).

Simplify to get \(y = -3x + 15\).

(ii) Differentiate \(f(x)\): \(f'(x) = 3x^2 - 18x + 24\).

Set \(f'(x) > 0\) for \(f(x)\) to be increasing:

\(3(x^2 - 6x + 8) > 0\)

Factorize: \(3(x-2)(x-4) > 0\)

The critical points are \(x = 2\) and \(x = 4\).

For \(f(x)\) to be increasing, \(x > 4\).

Thus, the smallest value of \(k\) is 4.