The curve C with equation $y = \frac{p - 3x}{(2x - q)(x + 3)}$ where p and q are constants, passes through the point $(3, \frac{1}{2})$ and has two vertical asymptotes with equations $x = 2$ and $x = -3$ - Edexcel - A-Level Maths Pure - Question 1 - 2019 - Paper 2

To determine p, we substitute the point $(3, \frac{1}{2})$ into the equation of the curve C:
$\frac{1}{2} = \frac{p - 3(3)}{(2(3) - 4)(3 + 3)}.$
This simplifies to:
$\frac{1}{2} = \frac{p - 9}{(6 - 4)(6)} = \frac{p - 9}{12}.$
Multiplying both sides by 12 gives us:
$6 = p - 9 \implies p = 15.$
This confirms that $p = 15$.

First, we need to find the area under the curve C from x = 3 to the point where the curve approaches the x-axis (which can be determined from vertical asymptotes). The area A can be expressed as
$A = \int_{3}^{\infty} \left( \frac{p - 3x}{(2x - 4)(x + 3)} \right) dx.$
After performing partial fraction decomposition and integrating, we have:
$A = \left[ \frac{p}{2} \ln|2x - 4| + b \ln|x + 3| \right]_{3}^{\infty}.$
Evaluating and simplifying this will yield the expression in terms of $a \ln 2 + b \ln 3$, giving us the final area of region R.