A curve $C$ has parametric equations
$x = 3t^2, \quad y = 6t^3, \quad \text{where } t \text{ is a real parameter.}$

1. Determine the domain, range, and Cartesian equation to establish the shape of the curve.

From $x = 3t^2$, since $t$ is a real parameter, $x \ge 0$.
Expressing $t^2$ in terms of $x$: $t^2 = \frac{x}{3}$.
Substituting into the square of $y$:
$y^2 = (6t^3)^2 = 36t^6 = 36\left(\frac{x}{3}\right)^3 = \frac{4}{3}x^3$
The curve exists only for $x \ge 0$ and is symmetric about the $x$-axis.

2. Determine the gradient at the origin and sketch the curve.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{18t^2}{6t} = 3t$
At the origin where $t=0$, $\frac{dy}{dx} = 0$, indicating a horizontal tangent (a cusp).
As $t \to \infty$, $x \to \infty$ and $y \to \infty$.
As $t \to -\infty$, $x \to \infty$ and $y \to -\infty$.
[Draw the axes, mark the origin, and sketch the two symmetric branches of the curve in the $x \ge 0$ region, getting steeper as $x$ increases.]

1. Differentiate $x$ and $y$ with respect to $t$ to find $\frac{dy}{dx}$.

$\frac{dx}{dt} = 6t$

$\frac{dy}{dt} = 18t^2$

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{18t^2}{6t} = 3t$

2. Equate the gradient of the tangent to the gradient of the given line to find $t$.

The gradient of the line $y = 4 - 2x$ is $-2$.

Since the tangent is parallel to this line, their gradients must be equal:
$3t = -2$
$t = -\frac{2}{3}$

3. Substitute $t$ into the parametric equations to find the exact coordinates of $P$.

$x = 3\left(-\frac{2}{3}\right)^2 = 3\left(\frac{4}{9}\right) = \frac{4}{3}$

$y = 6\left(-\frac{2}{3}\right)^3 = 6\left(-\frac{8}{27}\right) = -\frac{16}{9}$

The exact coordinates of $P$ are $\left(\frac{4}{3}, -\frac{16}{9}\right)$.

1. Find the derivative $\frac{dy}{dx}$ in terms of $t$ using the parametric equations.

$x = 3t^2 \implies \frac{dx}{dt} = 6t$
$y = 6t^3 \implies \frac{dy}{dt} = 18t^2$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{18t^2}{6t} = 3t$

2. Find the value of the parameter $t$ at the point $Q$.

At $Q\left(\frac{4}{3}, \frac{16}{9}\right)$, set $y = \frac{16}{9}$:
$6t^3 = \frac{16}{9}$
$t^3 = \frac{16}{54} = \frac{8}{27}$
$t = \frac{2}{3}$
(Check with $x$-coordinate: $3\left(\frac{2}{3}\right)^2 = \frac{4}{3}$, which is consistent).

3. Evaluate the gradient at $Q$ and formulate the equation of the tangent.

Gradient, $m = 3\left(\frac{2}{3}\right) = 2$

Equation of the tangent at $Q$:
$y - \frac{16}{9} = 2\left(x - \frac{4}{3}\right)$
Multiply by 9 to clear the fraction:
$9y - 16 = 18\left(x - \frac{4}{3}\right)$
$9y - 16 = 18x - 24$
$9y = 18x - 8$ (shown)

1. Find the coordinates of point $P$ based on the condition given in part (ii).

From the parametric equations, find $\frac{\mathrm{d}y}{\mathrm{d}x}$:
$\frac{\mathrm{d}x}{\mathrm{d}t} = 6t, \quad \frac{\mathrm{d}y}{\mathrm{d}t} = 18t^2 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{18t^2}{6t} = 3t$

The tangent at $P$ is parallel to $y = 4 - 2x$, which has a gradient of $-2$:
$3t = -2 \implies t = -\frac{2}{3}$

Substitute $t = -\frac{2}{3}$ into the equations for $C$:
$x = 3\left(-\frac{2}{3}\right)^2 = \frac{4}{3}, \quad y = 6\left(-\frac{2}{3}\right)^3 = -\frac{16}{9}$

So, $P$ has coordinates $\left(\frac{4}{3}, -\frac{16}{9}\right)$.

2. Find the coordinates of point $R$, the $x$-intercept of the tangent at $Q$.

The equation of the tangent at $Q$ is given as $9y = 18x - 8$.
To find where it cuts the $x$-axis, set $y = 0$:
$9(0) = 18x - 8$
$18x = 8 \implies x = \frac{8}{18} = \frac{4}{9}$

So, $R$ has coordinates $\left(\frac{4}{9}, 0\right)$.

3. Calculate the area of triangle $PQR$.

The coordinates are $P\left(\frac{4}{3}, -\frac{16}{9}\right)$, $Q\left(\frac{4}{3}, \frac{16}{9}\right)$, and $R\left(\frac{4}{9}, 0\right)$.

Notice that $P$ and $Q$ share the same $x$-coordinate ($\frac{4}{3}$), so $PQ$ forms a vertical line segment.
The length of the base $PQ$ is:
$\text{Base} = \frac{16}{9} - \left(-\frac{16}{9}\right) = \frac{32}{9}$

The perpendicular height from $R$ to the line $PQ$ is the horizontal distance between their $x$-coordinates:
$\text{Height} = \frac{4}{3} - \frac{4}{9} = \frac{8}{9}$

$\text{Area of } \triangle PQR = \frac{1}{2} \times \text{Base} \times \text{Height}$
$\text{Area} = \frac{1}{2} \times \frac{32}{9} \times \frac{8}{9} = \frac{128}{81}$

1. Substitute the parametric equations of $C$ into the equation of the tangent at $Q$.

The equation of the tangent at $Q$ is given as $9y = 18x - 8$.
Substitute $x = 3t^2$ and $y = 6t^3$ into the equation:
$9(6t^3) = 18(3t^2) - 8$
$54t^3 = 54t^2 - 8$
$54t^3 - 54t^2 + 8 = 0$
Divide by 2 to simplify:
$27t^3 - 27t^2 + 4 = 0$

2. Solve the cubic equation for $t$ by using the known repeated root at the point of tangency.

Find the parameter $t$ at the point $Q\left(\frac{4}{3}, \frac{16}{9}\right)$:
$3t^2 = \frac{4}{3} \implies t^2 = \frac{4}{9} \implies t = \frac{2}{3} \text{ (since } y > 0 \text{)}$
Because the line is tangent to $C$ at $Q$, $t = \frac{2}{3}$ must be a repeated root of the cubic equation. This means $(3t - 2)^2$ is a factor.
$(3t - 2)^2 = 9t^2 - 12t + 4$
By comparing coefficients (or long division), we factorise the cubic:
$27t^3 - 27t^2 + 4 = (9t^2 - 12t + 4)(3t + 1) = 0$
For point $S$, which is the other intersection point, we solve the remaining factor:
$3t + 1 = 0 \implies t = -\frac{1}{3}$

3. Find the coordinates of $S$ by substituting the new value of $t$ back into the parametric equations.

Substitute $t = -\frac{1}{3}$ into the parametric equations for $C$:
$x = 3\left(-\frac{1}{3}\right)^2 = 3\left(\frac{1}{9}\right) = \frac{1}{3}$
$y = 6\left(-\frac{1}{3}\right)^3 = 6\left(-\frac{1}{27}\right) = -\frac{2}{9}$
The coordinates of $S$ are $\left(\frac{1}{3}, -\frac{2}{9}\right)$.