Solution:

A triangle has vertices $i \lt j \lt k$ where $j - i,$ $k - j,$ and $k - i$ are all prime. Since $k - i = (j - i) + (k - j)$ is a prime that is a sum of two primes, and the sum of two odd primes is even, one of the two smaller differences must equal $2.$ So the differences are $\{2, p\}$ in some order with $p$ and $p + 2$ both prime: the twin prime pairs with $p + 2 \le 19$ are $(3, 5),$ $(5, 7),$ $(11, 13),$ and $(17, 19).$

For each pair, the middle vertex can be at distance $2$ or at distance $p$ from the smallest, and the total span is $p + 2,$ so there are $2\bigl(20 - (p + 2)\bigr)$ triangles. This gives $2 \cdot 15 = 30,$ $2 \cdot 13 = 26,$ $2 \cdot 7 = 14,$ and $2 \cdot 1 = 2$ for the four pairs.

The total is $30 + 26 + 14 + 2 = 72.$