jmc

Firstly, the congruence can be simplified to $3(6x+1)\equiv 4(\mathrm{mod}p)⟹18x\equiv 1(\mathrm{mod}p)$. This is solvable for $x$ if and only if $18$ is invertible modulo $p$, meaning $\gcd (18,p)=1$. Since the prime factors of $18$ are $2,3$, these are precisely the prime moduli for which an $x$ cannot exist since then $\gcd (18,p)>1$. Thus, the desired number is $2+3=\boxed{5}$.