Without loss of generality the coins form the pattern shown below.

Let $G$, $W$ and $O$ denote the total worth in shillings of the green, white and orange coins in the triangle, respectively. The problem reduces to showing that $G$, $W$ and $O$ are all divisible by three. Now fix some green coin $g'$ in the triangle, and consider a subset of the lines parallel to the sides of the triangle, such that every third parallel line is included, and all lines passing through $g'$ are included; this is illustrated below.

Since the total worth of the coins on each line is divisible by $3$, so is the sum of these quantities. However these lines contain each white coin exactly once, each orange coin exactly once, and a subset $S$ of the green coins exactly $3$ times. We thus have
$$W + O + 3G' \equiv 0 \pmod{3}$$
where $G'$ denotes the total worth of the green coins lying in the set $S$. Therefore
$$W + O \equiv 0 \pmod{3}.$$

But summing the worth of the coins in all lines parallel to one side of the triangle, we obtain
$$G + W + O \equiv 0 \pmod{3}.$$
Subtraction of these two congruences yields
$$G \equiv 0 \pmod{3},$$
and application of the same reasoning proves that $W \equiv O \equiv 0 \pmod{3}$.