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 2 - 2019 - Paper 1

To find $p$, substitute the point $(3, \frac{1}{2})$ into the curve's equation.
Setting $y = \frac{1}{2}$ when $x = 3$, we have:
$\frac{1}{2} = \frac{p - 3(3)}{(2(3) - 4)(3 + 3)}$
This simplifies to:
$\frac{1}{2} = \frac{p - 9}{(6 - 4)(6)}$
$\frac{1}{2} = \frac{p - 9}{2 \cdot 6}$
Multiplying both sides by $12$ results in:
$6 = p - 9$
Thus, solving for $p$ gives:
p = 15.

To find the area of region $R$, set up the integral bounded by the curve and the line $x = 3$.
This is expressed as: $\text{Area} = \int_{0}^{3} y \, dx = \int_{0}^{3} \frac{p - 3x}{(2x - 4)(x + 3)} \, dx$ Substituting $p = 15$ gives: $\int_{0}^{3} \frac{15 - 3x}{(2x - 4)(x + 3)} \, dx$. Use partial fractions to decompose the integrand which leads to finding constants $A$ and $B$, so: $\frac{15 - 3x}{(2x - 4)(x + 3)} = \frac{A}{2x - 4} + \frac{B}{x + 3}$. By solving for $A$ and $B$, proceed to evaluate the integral from $0$ to $3$, leading to the exact area in the form of $a \ln 2 + b \ln 3$.