The volume \(V\) of the solid of revolution is given by:

\(V = \pi \int_0^1 \left( \frac{6}{5 - 2x} \right)^2 \, dx\)

First, simplify the integrand:

\(\left( \frac{6}{5 - 2x} \right)^2 = \frac{36}{(5 - 2x)^2}\)

Thus, the integral becomes:

\(V = \pi \int_0^1 \frac{36}{(5 - 2x)^2} \, dx\)

Let \(u = 5 - 2x\), then \(du = -2 \, dx\) or \(dx = -\frac{1}{2} \, du\).

Change the limits of integration: when \(x = 0\), \(u = 5\); when \(x = 1\), \(u = 3\).

Substitute and integrate:

\(V = \pi \int_5^3 \frac{36}{u^2} \left(-\frac{1}{2}\right) \, du\)

\(= -18\pi \int_5^3 \frac{1}{u^2} \, du\)

\(= -18\pi \left[ -\frac{1}{u} \right]_5^3\)

\(= 18\pi \left( \frac{1}{3} - \frac{1}{5} \right)\)

\(= 18\pi \left( \frac{5 - 3}{15} \right)\)

\(= 18\pi \left( \frac{2}{15} \right)\)

\(= \frac{36\pi}{15}\)

\(= \frac{12\pi}{5}\)

Thus, the volume obtained is \(\frac{12}{5} \pi\).