The curve $C$ has equation $y = 2 + \frac{3a}{x - a}$, where $a$ is a positive constant.

1. Express the curve $C$ in terms of the base function $\mathrm{f}(x) = \frac{1}{x}$.

Let $\mathrm{f}(x) = \frac{1}{x}$.
The equation of $C$ is $y = \frac{3a}{x-a} + 2$.
This can be written as $y = 3a\mathrm{f}(x-a) + 2$.

2. Identify the scaling transformation.

The factor of $3a$ corresponds to $y = 3a\mathrm{f}(x)$, which is a scaling parallel to the $y$-axis by a scale factor of $3a$.

3. Identify the translation and state the sequence.

The terms $(x-a)$ and $+ 2$ correspond to a translation of $a$ units in the positive $x$-direction and $2$ units in the positive $y$-direction.
This is equivalent to a translation by the vector $\begin{pmatrix} a \\ 2 \end{pmatrix}$.
(Note: To achieve $+2$ strictly at the end, the scaling must precede the $y$-translation. Thus, stating the scaling first followed by the full vector translation works perfectly).

1. Find the equations of the asymptotes.

For the original curve $y = 2 + \frac{3a}{x-a}$, as $x \to \pm\infty$, $\frac{3a}{x-a} \to 0$, giving a horizontal asymptote $y = 2$.
The denominator is zero when $x = a$, giving a vertical asymptote $x = a$.
Applying the modulus function $y = \left| 2 + \frac{3a}{x-a} \right|$, the asymptotes remain $y = |2| = 2$ and $x = a$.

2. Find the coordinates of the $x$- and $y$-intercepts.

At the $x$-intercept, $y = 0$:
$2 + \frac{3a}{x-a} = 0$
$\frac{3a}{x-a} = -2$
$-2(x - a) = 3a \implies -2x + 2a = 3a \implies x = -\frac{a}{2}$
The $x$-intercept is $(-\frac{a}{2}, 0)$.

At the $y$-intercept, $x = 0$:
$y = \left| 2 + \frac{3a}{-a} \right| = |2 - 3| = |-1| = 1$
The $y$-intercept is $(0, 1)$.

3. Deduce the shape of the curve to sketch it.

For $x > a$, $2 + \frac{3a}{x-a} > 0$, so the curve approaches $y=2$ from above and goes to $+\infty$ near $x=a$.
For $x < a$, the original function is negative between $x = -\frac{a}{2}$ and $x = a$. The modulus reflects this portion in the $x$-axis, so it turns upwards from $(-\frac{a}{2}, 0)$, passes through $(0, 1)$, and goes to $+\infty$ near $x=a$.
For $x < -\frac{a}{2}$, the original function is positive and approaches $y=2$ from below as $x \to -\infty$.

1. Set up the two cases to remove the modulus for the equation.

$2 + \frac{3}{x - 1} = 3 \quad \text{or} \quad 2 + \frac{3}{x - 1} = -3$

2. Solve the first case.

$\frac{3}{x - 1} = 1 \\ x - 1 = 3 \\ x = 4$

3. Solve the second case.

$\frac{3}{x - 1} = -5 \\ -5(x - 1) = 3 \\ -5x + 5 = 3 \\ 5x = 2 \\ x = \frac{2}{5}$

4. State the solution to the inequality based on the points found.

From the shape of the graph sketched in part (b) (with $a=1$), the curve approaches the horizontal asymptote $y = 2$ at the extremities, so it falls below $y = 3$ in the outer regions bounded by the solutions.
$x < \frac{2}{5} \quad \text{or} \quad x > 4$