I claim all solutions are $(p,q,r)=(2,3,7)$ (and permutations of thereof) with $a=2$. Let $p\le q\le r$ without loss of generality. If $p,q,r$ are all odd, then we find that LHS is congruent to 2 modulo 4, a contradiction. So, $p=2$, yielding \[ (q^2+4)(r^2+4) = 13\left(4\cdot 13^{a-1}+1\right). \]If $q>3$ then the LHS of the equation above is congruent to 1 modulo 3, whereas the RHS is 2, a contradiction. So, $q=3$ and $r^2+4 = 4\cdot 13^{a-1}+1$. If $a-1$ is odd, then we find $7\mid r$, forcing $r=7$. In this case, $a=2$ works. If $a-1$ is even, then setting $a-1=2t$, we arrive at \[ 3 = (2\cdot 13^t -r)(2\cdot 13^t +r), \]which is easily seen to have no solutions.

The answer is $a = 2$, and $(p,q,r)$ is a permutation of $(2,3,7)$. These clearly work. Now we show they are the only ones. Taking modulo $3$ and $4$ gives that one of the primes is $2$ and another is $3$. WLOG $q = 3, r = 2$. We have $13p^2 + 39 = 4 \cdot 13^a$. Thus, $p^2 + 3 = 4 \cdot 13^{a - 1}$. Now assume $p \ne 7$. If $a - 1$ is odd, then taking modulo $7$ gives that $p^2 \equiv 0 \pmod 7$, absurd. Therefore, $a - 1$ is even. Hene \[ 3 = 4 \cdot 13^{a - 1} - p^2 = \left( 2 \cdot 13^{ \frac{a - 1}{2}} - p \right) \left(2 \cdot 13^{\frac{a-1}{2}} + p \right) \]Both factors are positive integers, so must be at most $3$, meaning $p < 3$, so $p = 2$, which obviously does not work. Thus, $p = 7$ and we are done.