From the conditions of the problem, we have that
$$3p+10=(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2+(n+3)^2=6n^2+6n+19,$$
so, we have that
$$3p=6n^2+6n+9,$$
and
$$p=2n^2+2n+3=2n(n+1)+3.$$
If one of the numbers $n$ or $n+1$ is divisible with $3$, then we have a contradiction with the condition that $p$ is a prime number. So, $n=3k+1$. Then,
$$p=2(3k+1)(3k+1+1)+3=2(3k+1)(3k+2)+3=2(9k^2+9k+2)+3=18k(k+1)+7.$$
Since $k(k+1)$ is an even number, we have that $36|p-7$.