The sum of the first \(n\) terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

Substituting the given values, \(a = -12\) and \(d = 6\), we have:

\(S_n = \frac{n}{2} [-24 + (n-1)6]\)

We need \(S_n > 3000\):

\(\frac{n}{2} [-24 + 6n - 6] > 3000\)

\(\frac{n}{2} [6n - 30] > 3000\)

\(3n(2n - 5) > 3000\)

\(3n^2 - 15n - 3000 > 0\)

Solving the quadratic inequality \(3n^2 - 15n - 3000 = 0\), we find the roots:

\(n = 34.2 \text{ and } n = -29.2\)

Since \(n\) must be a positive integer, the least possible value of \(n\) is \(35\).