Let $\sum_{i \in \emptyset} a_i = 0$ for ease. Then, $$\sum_{k=0}^n (-1)^k s_k = \sum_{k=0}^n (-1)^k \sum_{\substack{B \subseteq [n] \\ |B| = k}} 1_{p \mid \sum_{i \in B} a_i} \equiv \sum_{k=0}^n (-1)^k \sum_{\substack{B \subseteq [n] \\ |B| = k}} \left[1-(\sum_{i \in B} a_i)^{p-1}\right] \pmod{p}.$$ Since $$\sum_{k=0}^n (-1)^k\sum_{\substack{B \subseteq [n] \\ |B| = k}} 1 = \sum_{k=0}^n (-1)^k {n \choose k} = (1+(-1))^n = 0,$$ it suffices to show $$\sum_{k=0}^n (-1)^k\sum_{\substack{B \subseteq [n] \\ |B| = k}} \sum_{(i_1,\dots,i_{p-1}) \in B^{p-1}} a_{i_1}\dots a_{i_{p-1}} = 0.$$ The left hand side is $$\sum_{k=0}^n (-1)^k\sum_{\substack{1 \le t \le p-1 \\ t \le k}} \sum_{\substack{(i_1,\dots,i_{p-1}) \in [n]^{p-1} \\ |\{i_1,\dots,i_{p-1}\}| = t}} a_1\dots a_{p-1} \sum_{\substack{B \subseteq [n] \\ |B| = k}} 1_{B \ni i_1,\dots,i_{p-1}}$$ $$= \sum_{k=0}^n (-1)^k\sum_{\substack{1 \le t \le p-1 \\ t \le k}} \sum_{\substack{(i_1,\dots,i_{p-1}) \in [n]^{p-1} \\ |\{i_1,\dots,i_{p-1}\}| = t}} a_{i_1}\dots a_{i_{p-1}} {n-t \choose k-t}$$ $$ = \sum_{1 \le t \le p-1} \sum_{\substack{(i_1,\dots,i_{p-1}) \in [n]^{p-1} \\ |\{i_1,\dots,i_{p-1}\}| = t}} a_{i_1}\dots a_{i_{p-1}} \sum_{t \le k \le n} (-1)^k{n-t \choose k-t}.$$ And luckily, the inner sum is $$\sum_{m=0}^{n-t} (-1)^{m+t}{n-t \choose m} = (-1)^t(1+(-1))^{n-t} = 0,$$ where the last equality used $n > t$, which is true since $t \le p-1 \le n-1 < n$.
First, observe that we should have $x\le12$. Otherwise:
$$x^3(y^3+z^3)\ge x^6+x^3(xyz)\ge 13^6+2197(xyz)\gt2012(xyz+2).$$
Therefor $x \in \{1,2,3,4,5,6,7,8,9,10,11,12\}$.
But $x$ cannot be $3, 5,6 , 7, 9, 10, 11$ or $12$ because:
$$x|2012(xyz+2) \implies x|2012 \times2 \implies x|8 \times 503.$$
On the other hand $x$ cannot be $4$ or $8$ either because, in this case, we should have:
$$4^3|4 \times 503(4yz+2) \ or \ {} 8^3|4 \times 503(8yz+2),$$
both of which are impossible.
Therefore $x$ is either $1$ or $2$, and we have two new equations:
$$y^3+z^3=2012(yz+2) \\ y^3+z^3=503(yz+1).$$
Now, note that if $503|y$ then we get $503|z$, hence $503|y+z.$
Let's assume $503 \not| \ y$. Hence $503 \not| \ z$. By the Fermat's theorem, we have:
$$503|z^{502} - y^{502};$$
and because of $503| y^3+z^3$ we conclude that $503| y^{501}+z^{501}$, and as a result $503|y^{502}+yz^{501}$. Therefore:
$$503|z^{502}+yz^{501}\implies 503|y+z \implies y+z=503k.$$
Now, we have two other new equations:
$$ (1) \ k(y-z)^2+(k-4)yz=8 \\ (2)\ k(y-z)^2+(k-1)yz=1.$$
In the first case, it is very easy to see that the equation has no solution. First notice that $k$ should be $2$, and then we have $(y+z)^2-5yz= 4$ while $y+z=503 \times 2$. This means that the equation has no solution.
For the second equation, it is clear that $k$ must be equal to either $1$ or $2$. If $k=2$ then $z=y=1$, which is impossible. Thus $k=1$, and the only solution is $(2,251,252).$
EDIT: This problem is a shortlist problem of IMO $2012$ and has an official solution.
Best Answer
If we continue from here: $\;\;\;xyz+12 \mid x+y+z-41\;\;\;$ or $\;\;\;xyz+12|x+y+z+49$.
We can assume $x\leq y\leq z$. We see that $x,y,z\notin \{0,2,3,4,6,8,9\}$ since $xyz+12$ is prime.
If $x>1$ then $x,y\geq 5$ and so $xyz+12\geq 25z+12$ so in
first case: $$25z+1\leq |x+y+z-41|\leq 3z+41\implies z\le 1$$ and thus no solution.
second case: $$25z+1\leq x+y+z+49\leq 3z+49\implies z\leq 2$$ so no solution again.
If $x=1$ then $\;\;\;yz+12 \mid y+z-40\;\;\;$ or $\;\;\;yz+12|y+z+50$.
If $x=-1$ then we proccede similary like in a previous case...