Let $$p_n$$ denote the total number of triangles formed by joining the vertices of an $$n$$-side regular polygon. If $$p_{n+1} - p_n = 66$$, then the sum of all distinct prime divisors of $$n$$ is :

For a convex regular polygon, any choice of three distinct vertices forms a triangle. Hence, the total number of triangles that can be drawn is simply the combination

$$p_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6}$$

The condition given in the question is

$$p_{n+1} - p_n = 66$$

Compute each term:

$$p_{n+1}= \binom{n+1}{3}= \frac{(n+1)n(n-1)}{6}$$
$$p_n = \binom{n}{3}= \frac{n(n-1)(n-2)}{6}$$

Subtracting, we get

$$p_{n+1}-p_n = \frac{(n+1)n(n-1) - n(n-1)(n-2)}{6}$$

Factor out the common term $$n(n-1)$$:

$$p_{n+1}-p_n = \frac{n(n-1)\big[(n+1)-(n-2)\big]}{6} = \frac{n(n-1)\times 3}{6} = \frac{n(n-1)}{2}$$

Set this equal to the given value 66:

$$\frac{n(n-1)}{2}=66 \quad\Longrightarrow\quad n(n-1)=132$$

This is a quadratic equation:

$$n^2-n-132=0$$

Factoring (or using the quadratic formula) gives

$$(n-12)(n+11)=0\quad\Longrightarrow\quad n=12\;(\text{positive solution})$$

The distinct prime factors of $$12$$ are $$2$$ and $$3$$. Their sum is

$$2+3 = 5$$

Therefore, the required sum is $$5$$.

Option C which is: 5