(a) To find the equation of the line \(AB\), first calculate the gradient:

\(\text{Gradient of } AB = \frac{2 - (-1)}{5 - 2} = 1\)

Using point \(A(5, 2)\), the equation of the line is:

\(y - 2 = 1(x - 5)\)

Simplifying gives:

\(y = x - 3\)

(b) To find the volume of revolution, calculate the integral of the area between the curve \(x = y^2 + 1\) and the line \(y = x - 3\) rotated about the \(y\)-axis:

For the curve:

\(\pi \int (y^2 + 1)^2 \, dy = \pi \int (y^4 + 2y^2 + 1) \, dy\)

Integrating gives:

\(\pi \left( \frac{y^5}{5} + \frac{2y^3}{3} + y \right)\)

For the line:

\(\pi \int (y + 3)^2 \, dy = \pi \int (y^2 + 6y + 9) \, dy\)

Integrating gives:

\(\pi \left( \frac{y^3}{3} + 3y^2 + 9y \right)\)

Apply limits from \(y = -1\) to \(y = 2\):

\(\pi \left( \frac{8}{3} + 12 + 18 \right) - \pi \left( -\frac{1}{3} + 3 - 9 \right)\)

Calculating gives:

\(\pi \left( 39 - 15 \frac{3}{5} \right)\)

Volume = \(\frac{117}{5} \pi\) or approximately 73.5