To find the coefficient of \(x^6\) in the expansion of \(\left( 2x^3 - \frac{1}{x^2} \right)^7\), we use the binomial theorem:

\(\left( a + b \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

Here, \(a = 2x^3\) and \(b = -\frac{1}{x^2}\), and \(n = 7\).

We need the term where the power of \(x\) is 6. The general term is:

\(T_k = \binom{7}{k} (2x^3)^{7-k} \left(-\frac{1}{x^2}\right)^k\)

Simplifying, we get:

\(T_k = \binom{7}{k} \cdot 2^{7-k} \cdot x^{3(7-k)} \cdot (-1)^k \cdot x^{-2k}\)

\(= \binom{7}{k} \cdot 2^{7-k} \cdot (-1)^k \cdot x^{21-3k-2k}\)

\(= \binom{7}{k} \cdot 2^{7-k} \cdot (-1)^k \cdot x^{21-5k}\)

We want \(21-5k = 6\), so:

\(21 - 5k = 6\)

\(5k = 15\)

\(k = 3\)

Substitute \(k = 3\) into the term:

\(T_3 = \binom{7}{3} \cdot 2^{4} \cdot (-1)^3 \cdot x^{6}\)

\(= 35 \cdot 16 \cdot (-1) \cdot x^{6}\)

\(= -560 \cdot x^{6}\)

Thus, the coefficient of \(x^6\) is \(-560\).