Problem: Let $n$ be a positive integer.
Consider
\[ S = \left\{ (x,y,z) \mid
  x,y,z \in \{ 0, 1, \dots, n\}, \;
  x+y+z > 0 \right\} \]
as a set of $(n+1)^3-1$ points in the three-dimensional space.
Determine the smallest possible number of planes,
the union of which contains $S$ but does not include $(0,0,0)$.

\end{enumerate}

Solution: The answer is $3n$.
Here are two examples of constructions with $3n$ planes:
\begin{itemize}
  \ii $x+y+z=i$ for $i=1,\dots,3n$.
  \ii $x=i$, $y=i$, $z=i$ for $i=1,\dots,n$.
\end{itemize}
Suppose for contradiction we have $N < 3n$ planes.
Let them be $a_i x + b_i y + c_i z + 1 = 0$, for $i = 1, \dots, N$.
Define the polynomials
\begin{align*}
  A(x,y,z) &= \prod_{i=1}^n (x-i) \prod_{i=1}^n (y-i) \prod_{i=1}^n (z-i) \\
  B(x,y,z) &= \prod_{i=1}^N \left( a_i x + b_i y + c_i z + 1 \right).
\end{align*}
Note that $A(0,0,0) = (-1)^n  (n!)^3 \neq 0$
and $B(0,0,0) =  1 \neq 0$,
but $A(x,y,z) = B(x,y,z) = 0$ for any $(x,y,z) \in S$.
Also, the coefficient of $x^n y^n z^n$ in $A$ is $1$,
while the coefficient of $x^n y^n z^n$ in $B$ is $0$.

Now, define
\[ P(x,y,z) \coloneq A(x,y,z) - \lambda  B(x,y,z). \]
where $\lambda = \frac{A(0,0,0)}{B(0,0,0)} = (-1)^{n} (n!)^3$.
We now have that
\begin{itemize}
  \ii $P(x,y,z) = 0$ for any $x,y,z \in \left\{ 0,1,\dots,n \right\}^3$.
  \ii But the coefficient of $x^n y^n z^n$ is $1$.
\end{itemize}
This is a contradiction to Alon's combinatorial nullstellensatz.