Prime Pickle

Opening the brackets we have:- $\color{#D61F06}{p^2}+\color{#EC7300}{q^2} +\color{#20A900}{3(p+q)}=\color{#69047E}{r^2}+\color{#3D99F6}{3r}$ .

Modding both sides with $3$ , (only concentrating on the remainder obtained when dividing by $3$ ).

[Every square is equivalent to $0$ or $1$ modulo $3$ , because every number is of the form $3k), (3k+1), (3k+2)$ .]

So we get (match colors for more clarity):-

$\color{#D61F06}{0,1}+\color{#EC7300}{0,1} +\color{#20A900}{0}=\color{#69047E}{0,1}+\color{#3D99F6}{0}$ .

Clearly, if both $p$ and $q$ are not divisible by $3$ , we have $r^2 \equiv 2$ (mod 3), which is against maths.

So at lease one of $(p,q)$ is divisible by $3$ . Let us assume first that $3|p$ , but $p$ is a prime, so $p=3$ .

Putting $p=3$ in equation, we get on simplification:-

$(r-q)(r+q+3)=18$ . Only possible values of $q,r$ where $q$ is prime and $r$ is natural are $(2,4)$ and $(7,8)$ , which can easily be obtained by writing $18$ as a product of naturals.

But it can be the other way round also,ie, $q=3$ , so checking all the permutations, we finally have $4$ solutions $(p,q,r)$ , which are $(2,3,4)$ , $(3,2,4)$ , $(3,7,8)$ , and $(7,3,8)$ .

So, finally we get the sum of all possible values of $p$ as $\boxed{12}$

-Satvik