ABC is a triangle. D is the point on BC such that 3BD = 2DC.

1. Convert the given condition into a ratio and express $\overrightarrow{BC}$.

Since $3BD=2DC$,
$BD:DC=2:3.$
Also,
$\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}=-\mathbf m+\mathbf n=\mathbf n-\mathbf m.$

2. Find $\overrightarrow{BD}$ and hence $\overrightarrow{AD}$.

Since $D$ divides $BC$ in the ratio $2:3$,
$\overrightarrow{BD}=\frac{2}{5}\overrightarrow{BC}=\frac{2}{5}(\mathbf n-\mathbf m).$
Hence
$\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}$
$=\mathbf m+\frac{2}{5}(\mathbf n-\mathbf m)$
$=\frac{3}{5}\mathbf m+\frac{2}{5}\mathbf n.$

1. Express \(\overrightarrow{AR}\) in terms of \(\mathbf m\), \(\mathbf n\), and \(k\) using the result from part (a).

From part (a),
$\overrightarrow{AD}=\frac{3\mathbf m+2\mathbf n}{5}.$
Since $\overrightarrow{AR}=k\overrightarrow{AD},$
$\overrightarrow{AR}=\frac{k(3\mathbf m+2\mathbf n)}{5}.$

2. Form \(\overrightarrow{BR}\) using a vector route from \(B\) to \(R\).

$\overrightarrow{BR}=\overrightarrow{BA}+\overrightarrow{AR}$
Since $\overrightarrow{AB}=\mathbf m,$
$\overrightarrow{BA}=-\mathbf m.$
Hence
$\overrightarrow{BR}=-\mathbf m+\frac{k(3\mathbf m+2\mathbf n)}{5}$
$=\left(\frac{3k}{5}-1\right)\mathbf m+\frac{2k}{5}\mathbf n.$

3. Use the parallel condition \(BR \parallel AC\) to determine \(k\).

Since $AC\parallel BR$ and $\overrightarrow{AC}=\mathbf n,$ (\overrightarrow{BR}) must be a multiple of (\mathbf n).

Therefore the coefficient of (\mathbf m) in (\overrightarrow{BR}) is zero:
$\frac{3k}{5}-1=0$
$\frac{3k}{5}=1$
$k=\frac{5}{3}.$

1. Use the position of R on AD produced to compare the relevant heights.

From part (b)(i),
$AR=\frac{5}{3}AD.$
Since $R$ is on $AD$ produced,
$RD=AR-AD=\frac{5}{3}AD-AD=\frac{2}{3}AD.$
Triangles $ABD$ and $RBD$ have the same base $BD$. Their perpendicular heights to $BD$ are proportional to $AD$ and $RD$ because $A,D,R$ are collinear.

Hence
$\text{area of }\triangle ABD : \text{area of }\triangle RBD=AD:RD=AD:\frac{2}{3}AD=3:2.$