By Vieta, the sum of the roots is $-10(\bmod p)$. However, since the three roots are less than $p/3$, it follows that the roots are $\left(p-a^{\prime}\right)/3,\left(p-b^{\prime}\right)/3,\left(p-c^{\prime}\right)/3$, where there are finitely many choices $a^{\prime}<b^{\prime}<c^{\prime}$. By pigeonhole, one choice, say $(u, v, w)$ must occur for infinitely many $p$. We then get that the roots of $P$ are $-u/3,-v/3$, and $-w/3$. Moreover, we must have that $u, v, w$ are all $1(\bmod 3)$ or all $2(\bmod 3)$, and by Vieta, we have $u+v+w=30$. The polynomial is then uniquely determined by $u, v, w$. Thus, it suffices to count triples $u<v<w$ of positive integers such that $u, v, w$ are all $1(\bmod 3)$ or all $2(\bmod 3)$ and that $u+v+w=30$. It's not very hard to list them all now. When $u, v, w \equiv 1(\bmod 3)$, there are 7 triples: $(1,4,25),(1,7,22),(1,10,19),(1,13,16),(4,7,19)$, $(4,10,16)$, and $(7,10,13)$. When $u, v, w \equiv 2(\bmod 3)$, there are 5 triples: $(2,5,23),(2,8,20),(2,11,17),(5,8,17)$, and $(5,11,14)$. Hence, the answer is $7+5=12$.