(i) $2 x+y=39; \therefore a=2, b=1, c=39$

(ii) $\frac{2057}{18}$ units $^2$

(iii) (a) $\frac{2057}{3}$ units $^2$;  (b) $x=\frac{1}{54}(y+3)^2+4$

(i) $\begin{aligned} x &=3 t^2+2 \\ \frac{\mathrm{d} x}{\mathrm{~d} t} &=6 t \\ y &=6 t-1 \\ \frac{\mathrm{d} y}{\mathrm{~d} t} &=6 \\ \frac{\mathrm{d} y}{\mathrm{~d} x} &=\frac{\mathrm{d} y}{\mathrm{~d} t} \times \frac{\mathrm{d} t}{\mathrm{~d} x} \\ &=6 \times \frac{1}{6 t} \\ & \\ \text { [B1] } \end{aligned}$

(ii) At the intersection point of the Normal and the Parametric curve,
$$
\begin{array}{ll}
y=-2 x+39 & -(1) \\
x=3 t^2+2 & -(2) \\
y=6 t-1 & -(3)
\end{array}
$$

Substitute (2), (3) into (1):
$$
\begin{aligned}
6 t-1 &=-2\left(3 t^2+2\right)+39 \\
6 t-1 &=-6 t^2-4+39 \\
6 t^2+6 t-36 &=0
\end{aligned}
$$
From GC,
[B I]
$$
\begin{array}{l}
t=2 \text { or } t=-3\left(\text { Rejected }, t \geqslant \frac{1}{6}\right) \\
x=3(2)^2+2=14 \\
y=6(2)-1=11
\end{array}
$$
[M1]
Required area
$$
\begin{array}{l}
=\int_{\frac{25}{12}}^{14} y_C \mathrm{~d} x+\int_{14}^{195} y_N \mathrm{~d} x \\
=\int_{\frac{1}{6}}^2(6 t-1)(6 t) \mathrm{d} t+\int_{14}^{19.5}(-2 x+39) \mathrm{d} x \\
=\int_{\frac{1}{6}}^2 36 t^2-6 t \mathrm{~d} t+30.25
\end{array}
$$
From GC,
$$
\begin{array}{l}
=\frac{3025}{36}+30.25 \\
=\frac{2057}{18} \text { units }^2
\end{array}
$$

(iii) (a) For scale factor of 2 in the $x$-direction, all the $x$-values multiply by 2 .
For scale factor of 3 in the $x$-direction, all the $y$-values multiply by 3 .
Required area
$$
\begin{array}{l}
=\frac{2057}{18} \times 2 \times 3 \\
=\frac{2057}{3} \mathrm{units}^2
\end{array}
$$

(b) $x=3 t^2+2$
$$
y=6 t-1
$$
After the 2 transformations,
$$
\frac{x}{2}=3 t^2+2 \Rightarrow x=6 t^2+4
$$
$\frac{y}{3}=6 t-1 \Rightarrow y=18 t-3 \Rightarrow t=\frac{y+3}{18} \quad$ [M1]
Cartesian equation of $D$ :
$$
\begin{array}{l}
x=6\left(\frac{y+3}{18}\right)^2+4 \\
x=\frac{1}{54}(y+3)^2+4
\end{array}
$$