To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Given \(y = \frac{6}{2x-3}\), we have:

\(y^2 = \left(\frac{6}{2x-3}\right)^2 = \frac{36}{(2x-3)^2}\)

The integral becomes:

\(V = \pi \int_{2}^{3} \frac{36}{(2x-3)^2} \, dx\)

Using the substitution \(u = 2x-3\), \(du = 2 \, dx\), \(dx = \frac{du}{2}\), the limits change from \(x = 2\) to \(u = 1\) and \(x = 3\) to \(u = 3\).

The integral becomes:

\(V = \pi \int_{1}^{3} \frac{36}{u^2} \cdot \frac{1}{2} \, du = 18\pi \int_{1}^{3} u^{-2} \, du\)

Integrating, we get:

\(18\pi \left[ -u^{-1} \right]_{1}^{3} = 18\pi \left( -\frac{1}{3} + 1 \right) = 18\pi \left( \frac{2}{3} \right) = 12\pi\)

Thus, the volume is \(12\pi\).