Always prime

I first noticed that $117$ is divisible by $3$ because of the digit sum, so I decided to consider the congruence mod $3$ . We have that $p^4 + q^4 + r^4 + 117 \equiv p^2 + q^2 + r^2 \mod 3$ . Remember that all primes distinct from $3$ are either of the form $3k + 1$ or $3k - 1$ which implies that if none of the primes are $3$ then this congruence will evaluate to $1 + 1 + 1 \equiv 0 \mod 3$ . This means that $p^4 + q^4 + r^4 + 117$ won't be a prime number, as it is divisible by 3 (and trivially, it must be much bigger than 3). This implies that one of the primes must be $3$ . Notice that in the expression we actually have perfect symmetry so without loss of generality let's assume that $p = 3$ . We now have a different problem:

Find primes $q,r$ such that $q^4 + r^4 + 198$ is prime. We now notice that $198$ is even so let's consider the congruence mod $2$ . $q^4 + r^4 + 198 \equiv q^4 + r^4 \mod 2$ . If neither of the primes are equal to $2$ , then $q^4 + r^4 \equiv 1 + 1 \equiv 0 \mod 2$ . This means that one of the primes must be $2$ , so let $q = 2$ .

Finally, we have to find a prime number $r$ such that $r^4 + 214$ is prime. Now remember that if $r \neq 5$ we have $r^4 \equiv 1 \mod 5$ which would imply $r^4 + 214 \equiv 1 + 214 \equiv 215 \equiv 0 \mod 5$ so it wouldn't be prime. Therefore, necessarily $r = 5$ .

It is hard to check but $2^4 + 3^4 + 5^4 + 117 = 839$ is indeed a prime number. Finally, we want to maximize the linear expression. Let's simply assign the biggest primes to the biggest coefficients and compute $15 \times 5 + 13 \times 3 + 11 \times 2 = 136$ .