(i) Since \(P(3, -10)\) is a stationary point, \(f'(3) = 0\). Substituting into \(f'(x) = 2x^2 + kx - 12\), we get:

\(2(3)^2 + 3k - 12 = 0\)

\(18 + 3k - 12 = 0\)

\(3k = -6\)

\(k = -2\)

To find the other stationary point, set \(f'(x) = 0\):

\(2x^2 - 2x - 12 = 0\)

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

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

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

Thus, the \(x\)-coordinate of \(Q\) is \(-2\).

(ii) Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = 4x - 2\)

At \(x = 3\), \(f''(3) = 4(3) - 2 = 10 > 0\), so \(P\) is a minimum.

At \(x = -2\), \(f''(-2) = 4(-2) - 2 = -10 < 0\), so \(Q\) is a maximum.

(iii) Integrate \(f'(x)\) to find \(f(x)\):

\(f(x) = \int (2x^2 - 2x - 12) \, dx\)

\(f(x) = \frac{2}{3}x^3 - x^2 - 12x + c\)

Using \(P(3, -10)\), substitute to find \(c\):

\(-10 = \frac{2}{3}(3)^3 - (3)^2 - 12(3) + c\)

\(-10 = 18 - 9 - 36 + c\)

\(c = 17\)

Thus, \(f(x) = \frac{2}{3}x^3 - x^2 - 12x + 17\).