(i) To express \(f(x) = 2x^2 - 12x + 7\) in the form \(a(x-b)^2 - c\), we complete the square:

First, factor out the 2 from the quadratic terms: \(2(x^2 - 6x) + 7\).

Complete the square inside the parentheses: \(x^2 - 6x = (x-3)^2 - 9\).

Thus, \(2(x^2 - 6x) = 2((x-3)^2 - 9) = 2(x-3)^2 - 18\).

Adding the constant term: \(2(x-3)^2 - 18 + 7 = 2(x-3)^2 - 11\).

So, \(f(x) = 2(x-3)^2 - 11\).

(ii) The range of \(f\) is determined by the vertex form \(2(x-3)^2 - 11\). The minimum value occurs at \(x = 3\), giving \(f(3) = -11\). Therefore, the range is \(f \geq -11\).

(iii) To find the set of values of \(x\) for which \(f(x) < 21\), solve:

\(2x^2 - 12x + 7 < 21\).

Rearrange: \(2x^2 - 12x + 7 - 21 < 0\) or \(2x^2 - 12x - 14 < 0\).

Complete the square: \(2(x-3)^2 - 11 < 21\).

Rearrange: \(2(x-3)^2 < 32\).

Divide by 2: \((x-3)^2 < 16\).

Take the square root: \(-4 < x-3 < 4\).

Thus, \(-1 < x < 7\).