Thanks Guest,
Ok lets look at this. I am redoing your working in the hope that all becomes clear to me.
First a definition:
Prime triplet .
set of three prime numbers Which form of arithmetic sequence with common difference two is called a triplet prime .
Reference: https://artofproblemsolving.com/wiki/index.php/Prime_triplet
That defintiion appears to be nonesense because, this site https://en.wikipedia.org/wiki/Prime_triplet
says that 7,11 and 13 are prime triplets.
I assume prime triplets are any 3 consecutively prime numbers .... Is that correct? For now I will assume so.
\(\color{blue}{\text{We consider all positive prime triplets }P
We have
P, Q, R all consecutive primes
If P=2 then Q=3 and R=5 4+9+25=38 which is not prime so P is bigger than 2
So, let
\(P=3k_1+a\\ Q=3k_2+b\\ R=3k_3+c\\\)
where a,b and c are can equal 0,1 or 2
\(P^2+Q^2+R^2\\ =(3k_1+a)^2+(3k_2+b)^2+(3k_3+c)^2\\ =9k_1^2+6ak+a^2+9k_2^2+6bk+b^2+9k_3^2+6ck+c^2\\ =9k_1^2+9k_2^2+9k_3^2+6ak+6bk+6ck+a^2+b^2+c^2\\ =3(3k_1^2+3k_2^2+3k_3^2+2ak+2bk+2ck)+a^2+b^2+c^2\\ \)
Ok now please explain why this must be a multiple of 3 .....