Solution:

Let $P$ be a good point. Let $l, m, n$ be respectively the number of parts the sides $BC, CA, AB$ are divided by the rays starting from $P$. Note that a ray must pass through each of the vertices of the triangle $ABC$; otherwise we get some quadrilaterals.

Let $h_1$ be the distance of $P$ from $BC$. Then $h_1$ is the height for all the triangles with their bases on $BC$. Equality of areas implies that all these bases have equal length. If we denote this by $x$, we get $l x = a$. Similarly, taking $y$ and $z$ as the lengths of the bases of triangles on $CA$ and $AB$ respectively, we get $m y = b$ and $n z = c$. Let $h_2$ and $h_3$ be the distances of $P$ from $CA$ and $AB$ respectively. Then

$$h_1 x = h_2 y = h_3 z = \frac{2 \Delta}{27}$$

where $\Delta$ denotes the area of the triangle $ABC$. These lead to

$$h_1 = \frac{2 \Delta}{27} \frac{l}{a}, \quad h_2 = \frac{2 \Delta}{27} \frac{m}{b}, \quad h_3 = \frac{2 \Delta}{27} \frac{n}{c}$$

But

$$\frac{2 \Delta}{a} = h_a, \quad \frac{2 \Delta}{b} = h_b, \quad \frac{2 \Delta}{c} = h_c$$

Thus we get

$$\frac{h_1}{h_a} = \frac{l}{27}, \quad \frac{h_2}{h_b} = \frac{m}{27}, \quad \frac{h_3}{h_c} = \frac{n}{27}$$

However, we also have

$$\frac{h_1}{h_a} = \frac{[PBC]}{\Delta}, \quad \frac{h_2}{h_b} = \frac{[PCA]}{\Delta}, \quad \frac{h_3}{h_c} = \frac{[PAB]}{\Delta}$$

Adding these three relations,

$$\frac{h_1}{h_a} + \frac{h_2}{h_b} + \frac{h_3}{h_c} = 1$$

Thus

$$\frac{l}{27} + \frac{m}{27} + \frac{n}{27} = \frac{h_1}{h_a} + \frac{h_2}{h_b} + \frac{h_3}{h_c} = 1$$

We conclude that $l + m + n = 27$. Thus every good point $P$ determines a partition $(l, m, n)$ of $27$ such that there are $l, m, n$ equal segments respectively on $BC, CA, AB$.

Conversely, take any partition $(l, m, n)$ of $27$. Divide $BC, CA, AB$ respectively into $l, m, n$ equal parts. Define

$$h_1 = \frac{2 l \Delta}{27 a}, \quad h_2 = \frac{2 m \Delta}{27 b}$$

Draw a line parallel to $BC$ at a distance $h_1$ from $BC$; draw another line parallel to $CA$ at a distance $h_2$ from $CA$. Both lines are drawn such that they intersect at a point $P$ inside the triangle $ABC$. Then

$$[PBC] = \frac{1}{2} a h_1 = \frac{l \Delta}{27}, \quad [PCA] = \frac{m \Delta}{27}$$

Hence

$$[PAB] = \frac{n \Delta}{27}$$

This shows that the distance of $P$ from $AB$ is

$$h_3 = \frac{2 n \Delta}{27 c}$$

Therefore each triangle with base on $CA$ has area $\frac{\Delta}{27}$. We conclude that all the triangles which partition $ABC$ have equal areas. Hence $P$ is a good point.

Thus the number of good points is equal to the number of positive integral solutions of the equation $l + m + n = 27$. This is equal to

$$\binom{26}{2} = 325$$