Subtracting the second equation from the first one gives $3p + 2q - 3s = 0$ which implies $2q = 3(s - p)$. Thus $2q$ is divisible by $3$. As both $2$ and $q$ are primes, this implies $q = 3$. Substituting $q = 3$ into the initial system of equations and simplifying gives
$$\begin{cases} 6p + 5r + 3s = 115, \\ 3p + 5r + 6s = 121. \end{cases} \quad (1)$$
If $r$ and $s$ were both odd then also $5r$ and $3s$ would be odd, whence $5r + 3s$ would be even. As $6p$ is even, too, the left-hand side of the first equation of system (1) would be even and could not equal the right-hand side $115$. Thus one of $r$ and $s$ is even, i.e., $r = 2$ or $s = 2$. Analogously if $p$ and $r$ were both odd then the second equation of system (1) would give a contradiction; hence $p = 2$ or $r = 2$. Consequently, if $r \neq 2$ then $p = s = 2$, but substituting $p = s$ into system (1) and subtracting the second equation from the first one gives $0 = -6$. The contradiction shows that $r = 2$. Substituting $r = 2$ into system (1) and simplifying gives
$$\begin{cases} 6p + 3s = 105, \\ 3p + 6s = 111. \end{cases}$$
Solving this equation gives $p = 11$ and $s = 13$.

As in Solution 1 we find $q = 3$ and obtain system (1). Subtracting the first equation from the second one in system (1) gives $3s - 3p = 6$, implying $s = p + 2$. The least pairs of prime numbers with difference $2$ are $(3,5)$, $(5,7)$ and $(11,13)$. The next such pair $(17,19)$ gives $6p + 3s > 115$, implying that no more pairs are suitable. Substituting the three pairs one by one into system (1) and simplifying each time gives $5r = 82$, $5r = 64$ and $5r = 10$, respectively. Only the last alternative leads to an integral solution $r = 2$, which is a prime, too. Consequently, the only possibility is $(p, q, r, s) = (11, 3, 2, 13)$.