Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of $f(x)=3 x+2$ and below by the $x$-axis over the interval $[2,3]$ about the line $y=-2$. Enter an exact value in terms of $\pi$.

Step 1 ：We are given the function \(f(x) = 3x + 2\) and we are asked to find the volume of the solid formed by revolving the region bounded by the graph of \(f(x)\), the x-axis, and the lines \(x = 2\) and \(x = 3\) about the line \(y = -2\).

Step 2 ：We use the washer method to find the volume of the solid of revolution. The formula for the volume of a washer is \(\pi \int_{a}^{b} [R(x)]^2 - [r(x)]^2 dx\), where \(R(x)\) is the outer radius and \(r(x)\) is the inner radius.

Step 3 ：In this case, the outer radius \(R(x)\) is the distance from the line \(y = -2\) to the graph of \(f(x)\), which is \(f(x) - (-2) = 3x + 4\). The inner radius \(r(x)\) is the distance from the line \(y = -2\) to the x-axis, which is \(-2\).

Step 4 ：The limits of integration are \(a = 2\) and \(b = 3\).

Step 5 ：Substituting these values into the formula, we get \(\pi \int_{2}^{3} [(3x + 4)]^2 - (-2)^2 dx\).

Step 6 ：Evaluating this integral gives us the volume of the solid of revolution, which is \(129\pi\).

Step 7 ：Final Answer: The volume of the solid of revolution is \(\boxed{129\pi}\).