Let the first term be \(a\) and the common ratio be \(r\). Then the \(n\)-th term is \(T_n = ar^{n-1}\).

We are given:

\(T_5 = ar^4 = 8k,\quad T_6 = ar^5 = -12,\quad T_7 = ar^6 = 2k.\)

1. Find \(k\) and \(r\)

In a geometric progression, \[ r = \frac{T_6}{T_5} = \frac{-12}{8k} = -\frac{3}{2k}, \qquad r = \frac{T_7}{T_6} = \frac{2k}{-12} = -\frac{k}{6}. \] Equating these: \[ -\frac{3}{2k} = -\frac{k}{6} \;\Rightarrow\; \frac{3}{2k} = \frac{k}{6} \;\Rightarrow\; 3 \cdot 6 = 2k \cdot k \;\Rightarrow\; 18 = 2k^2 \;\Rightarrow\; k^2 = 9. \] So \(k = \pm 3\). Since \(k\) is negative, \(k = -3\).

Now \[ r = -\frac{k}{6} = -\frac{-3}{6} = \frac{1}{2}. \] As \(|r| = \tfrac{1}{2} < 1\), the sum to infinity exists.

2. Find the first term \(a\)

From \(T_5 = ar^4 = 8k\), \[ a = \frac{8k}{r^4}. \] Since \(r = \tfrac{1}{2}\), we have \(r^4 = \left(\tfrac{1}{2}\right)^4 = \tfrac{1}{16}\), so \[ a = \frac{8k}{1/16} = 8k \cdot 16 = 128k. \] With \(k = -3\), \[ a = 128(-3) = -384. \]

3. Sum to infinity

For \(|r| < 1\), the sum to infinity is \[ S_\infty = \frac{a}{1 - r} = \frac{-384}{1 - \tfrac{1}{2}} = \frac{-384}{\tfrac{1}{2}} = -384 \times 2 = -768. \]

\(\boxed{S_\infty = -768}\)