[Proof] If
$$2 a b c-b c-c a-a b=x b c+y c a+z a b,$$

where $x \geqslant 0, y \geqslant 0, z \geqslant 0$, then we have
$$2 a b c=b c(x+1)+c a(y+1)+a b(z+1).$$

Thus $a \mid (x+1) b c$ and since $(a, b c)=1$, it follows that $a \mid (x+1)$. Therefore, $a \leqslant x+1$.
Similarly, we have $b \leqslant y+1, c \leqslant z+1$, hence
$$\begin{aligned} 3 a b c & =b c a+c a b+a b c \\ & \leqslant b c(x+1)+c a(y+1)+a b(z+1) \\ & =2 a b c, \text{ a contradiction. } \end{aligned}$$

Therefore, $2 a b c-b c-c a-a b$ cannot be expressed in the form $x b c+y c a+z a b$.
Next, to prove that any number $n$ greater than $2 a b c-b c-c a-a b$ can be expressed in the form $x b c+y c a+z a b$, we will prove a simple result to be used below:

Let $a, b$ be positive integers and $(a, b)=1$, then any number greater than $a b-a-b$ can be expressed in the form $a x+b y (x \geqslant 0, y \geqslant 0)$.

Since $(a, b)=1$, every integer $m$ can be expressed as
$$m=u a+v b, u, v \in \mathbb{Z}.$$

Clearly, such an expression is not unique, but all expressions can be written as
$$m=(u-k b) a+(v+k a) b, k \in \mathbb{Z}.$$

Choose $k_0$ such that $0 \leqslant u-k_0 b < a$. Let $u_0 = u - k_0 b$ and $v_0 = v + k_0 a$. Then $0 \leqslant u_0 < a$ and $m = u_0 a + v_0 b$.

Since $m > a b - a - b$, we have
$$u_0 a + v_0 b > a b - a - b.$$

This implies $v_0 b > -b$, so $v_0 > -1$, i.e., $v_0 \geqslant 0$. This proves the above proposition.
Finally, we use the above proposition to prove that when $n > 2 a b c - b c - c a - a b$, $n$ can be expressed in the form $x b c + y c a + z a b (x \geqslant 0, y \geqslant 0, z \geqslant 0)$.
Since
$$\begin{aligned} n & > 2 a b c - b c - c a - a b \\ & = (a b c - a b - c a + a) + (a b c - a - b c) \\ & = a (b - 1) (c - 1) + (a b c - a - b c) \\ & \geqslant a b c - a - b c. \end{aligned}$$

And since $(a, b c) = 1$, $n$ can be written as $n = a w + x b c$ with $x \leqslant a - 1$. Thus,
$$a w = n - b c x \geqslant n - b c (a - 1) > a b c - a b - c a,$$

i.e., $w > b c - b - c$. Since $(b, c) = 1$, we have $w = c y + b z (y \geqslant 0, z \geqslant 0)$. Therefore,
$$\begin{aligned} n & = a w + b c x \\ & = b c x + c a y + a b z. \end{aligned}$$