ORS 是一个三角形。 J 是 RS 上的一个点，使得 $RJ:JS = 3:2$。 $\overrightarrow{OR} = r$ 且 $\overrightarrow{OS} = s$。

1. Express the vector from $R$ to $S$ and identify the fraction of $RS$ from $R$ to $J$.

Since $RJ:JS=3:2$, the whole line segment $RS$ is split into $5$ equal parts, and
$\\overrightarrow{RJ}=\\frac{3}{5}\\overrightarrow{RS}.$
Also,
$\\overrightarrow{RS}=\\overrightarrow{OS}-\\overrightarrow{OR}=s-r.$

2. Find $\\overrightarrow{OJ}$ by following the path $O \to R \to J$ and simplify.

$\\overrightarrow{OJ}=\\overrightarrow{OR}+\\overrightarrow{RJ}$
$=r+\\frac{3}{5}(s-r)$
$=r+\\frac{3}{5}s-\\frac{3}{5}r$
$=\\frac{2}{5}r+\\frac{3}{5}s$
$=\\frac{1}{5}(2r+3s).$

1. Find the position vector of $X$ from $O$.

$\overrightarrow{OX}=\overrightarrow{OR}+\overrightarrow{RX}$
$=r+\frac{1}{5}(r+9s)$
$=\frac{1}{5}(6r+9s)$

2. Compare $\overrightarrow{OX}$ with the given result for $\overrightarrow{OJ}$ and conclude collinearity.

From part (a),
$\overrightarrow{OJ}=\frac{1}{5}(2r+3s).$
Now,
$\overrightarrow{OX}=\frac{1}{5}(6r+9s)=\frac{3}{5}(2r+3s)=3\overrightarrow{OJ}.$
Hence $\overrightarrow{OX}$ is a scalar multiple of $\overrightarrow{OJ}$.