Step 1: Understanding the Concept:
We find the general term of the expansion and determine for which value of \(r\) the power of \(x\) is a natural number.
Step 2: Detailed Explanation:
The general term is:
\[ T_{r+1} = \binom{9}{r} (2x)^{1/2(9-r)} (3x^{-2})^{r/3} \]
\[ T_{r+1} = \binom{9}{r} 2^{(9-r)/2} 3^{r/3} x^{(9-r)/2 - 2r/3} \]
Power of \(x\):
\[ P = \frac{27 - 3r - 4r}{6} = \frac{27 - 7r}{6} \]
For the coefficient to be \(k \neq 0\), \(P\) must be an integer \(m \in \mathbb{N}\).
Test values of \(r \in \{0, 1, \dots, 9\}\):
- If \(r = 3\), \(P = \frac{27 - 21}{6} = \frac{6}{6} = 1\). This satisfies the condition.
Substituting \(r = 3\):
\[ k = \binom{9}{3} 2^{(9-3)/2} 3^{3/3} \]
\[ k = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 2^{3} \times 3 \]
\[ k = 84 \times 8 \times 3 = 2016 \]
Step 3: Final Answer:
The value of \(k\) is 2016.