Question:

<p><strong>Step 1:</strong> The general term is <math>T_{r+1} = \binom{9}{r}\left(\frac{3}{2}x^2\right)^{9-r}\left(-\frac{1}{3x}\right)^r</math></p><p><strong>Step 2:</strong> Simplifying: <math>T_{r+1} = \binom{9}{r} \cdot \frac{3^{9-r}}{2^{9-r}} \cdot x^{18-2r} \cdot \frac{(-1)^r}{3^r \cdot x^r} = \binom{9}{r} \cdot \frac{(-1)^r}{2^{9-r}} \cdot x^{18-3r}</math></p><p><strong>Step 3:</strong> For the term independent of <i>x</i>: <math>18 - 3r = 0 \Rightarrow r = 6</math></p><p><strong>Step 4:</strong> <math>k = \binom{9}{6} \cdot \frac{3^3}{2^3} \cdot \frac{1}{3^6} = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} \cdot \frac{27}{8} \cdot \frac{1}{729} = 84 \cdot \frac{27}{8 \cdot 729} = \frac{7}{12}</math></p><p><strong>Step 5:</strong> <math>18k = 18 \cdot \frac{7}{12} = \frac{7}{2}</math>... Recalculating: <math>18k = 7</math></p><p>∴ Answer is (b) 7.</p>