Example 3 Let $n$ be a positive integer,
$$
\begin{aligned}
S= & \{(x, y, z) \mid x, y, z \in\{0,1, \cdots, n\}, \\
& x+y+z>0\}
\end{aligned}
$$

is a set of $(n+1)^{3}-1$ points in three-dimensional space. Try to find the minimum number of planes whose union contains $S$ but does not contain $(0,0,0)$. ${ }^{[4]}$
Suppose there exist $m$ planes
$$
a_{i} x+b_{i} y+c_{i} z-d_{i}=0\left(1 \leqslant i \leqslant m, d_{i} \neq 0\right)
$$

satisfying the conditions.
Construct the $m$-degree polynomial
$$
f(x, y, z)=\prod_{i=1}^{m}\left(a_{i} x+b_{i} y+c_{i} z-d_{i}\right) \text {. }
$$

By the problem, when $(x, y, z) \in S$,
$$
f(x, y, z) \equiv 0 \text {, but } f(0,0,0) \neq 0 \text {. }
$$

If $m<3 n$, by Theorem 4 (3) it is easy to get
$$
\Delta_{x}^{n} \Delta_{y}^{n} \Delta_{z}^{n} f \equiv 0 .
$$

Also, by Theorem 3 (2) we have
$$
\Delta^{n} f(x)=\sum_{i=0}^{n}(-1)^{n-i} \mathrm{C}_{n}^{i} f(x+i) \text {. }
$$

Generalizing to three dimensions, we have
$$
\begin{array}{l}
\Delta_{x}^{n} \Delta_{y}^{n} \Delta_{z}^{n} f \\
=\sum_{(i, j, k) \in S U \mid(0,0,0)!}(-1)^{3 n-i-j-k} \mathrm{C}_{n}^{i} \mathrm{C}_{n}^{j} \mathrm{C}_{n}^{k} \text {. } \\
f(x+i, y+j, z+k) \text {. } \\
\end{array}
$$

Let $x=y=z=0$, we get
$$
\begin{array}{l}
f(0,0,0) \\
=\sum_{(i, k, k) \in s}(-1)^{i+j+k+1} \mathrm{C}_{n}^{i} \mathrm{C}_{n}^{j} \mathrm{C}_{n}^{k} f(i, j, k)=0 \text {. } \\
\end{array}
$$

This contradicts the problem's assumption.
Therefore, $\dot{m} \geqslant 3 n$.
Construct $3 n$ planes to include all points in $S$.
Notice that each point in $S$ has at least one coordinate as $1,2, \cdots, n$.
Thus, $3 n$ planes
$$
\begin{array}{l}
x-i=0(1 \leqslant i \leqslant n), \\
y-j=0(1 \leqslant j \leqslant n), \\
z-k=0(1 \leqslant k \leqslant n)
\end{array}
$$

clearly meet the requirements.
Hence, the minimum number of planes that satisfy the conditions is $3 n$.