Denote $f(n) = 2n^3 + 4n^2 - 3n + 12$. The following table shows the remainders of $n^2, n^3, n^5$ and $f(n)$ upon division by 11:

As one can see from the table, the only remainders upon division by 11 that the fifth power of an arbitrary integer $n$ can give are 0, 1 and 10. On the other hand, integers of the form $f(n)$ give only remainders 1, 3, 4, 5, 6, 7, and 9 upon division by 11, whereby the remainder is 1 only if $n$ is divisible by 11. Consequently, $f(p)$ can be the fifth power of an integer only if $p$ is divisible by 11. As $p$ is prime, the only possibility is $p = 11$. And indeed, $f(11) = 2 \cdot 11^3 + 4 \cdot 11^2 - 3 \cdot 11 + 12 = 3125 = 5^5$.