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:

The number of triangles formed by joining the vertices of an $n$-sided regular polygon is given by $p_n = ^{n}C_{3}$.

Given that $p_{n+1} - p_n = 66$.

$^{n+1}C_{3} - ^{n}C_{3} = 66$

Using the property $^{n+1}C_{r} - ^{n}C_{r} = ^{n}C_{r-1}$, we get:

$^{n}C_{2} = 66$

$\dfrac{n(n-1)}{2} = 66$

$n(n-1) = 132$

$n^2 - n - 132 = 0$

$(n-12)(n+11) = 0$

Since $n$ must be a positive integer, $n = 12$.

The prime factorization of $12$ is $2^2 \times 3$. The distinct prime divisors of $12$ are $2$ and $3$.

The sum of all distinct prime divisors of $n$ is $2 + 3 = 5$.

Answer: $5$