首先，我们需要利用给定的代数关系，化简出变量之间的范围限制： 将 $c = b + 2$ 代入 $a + b + c = 10$ 中：

因为 $a, b, c$ 都是大于 0 的正整数（即 $a, b, c \geqslant 1$）：

• 由 $b \geqslant 1$ 得 $2b \geqslant 2$，从而 $a = 8 - 2b \leqslant 6$。
• 由 $a \geqslant 1$ 得 $8 - 2b \geqslant 1 \implies 2b \leqslant 7 \implies b \leqslant 3.5$。
• 由于 $b$ 必须是整数，所以 $b$ 只能取 $1, 2, 3$。

因此，本题只有 3 种可能的整数情况。

根据上述化简，穷举所有可能的正整数三元组 $(a, b, c)$，计算它们的乘积 $abc$，并验证每个乘积是否为偶数。

Substitute $c = b + 2$ into $a + b + c = 10$:

Since $a$, $b$, and $c$ are integers greater than 0, we must have $b \geqslant 1$. Also:

Since $b$ must be an integer, the only possible values for $b$ are $1$, $2$, and $3$. We list the cases:

1. Case 1: $b = 1$ $a = 8 - 2(1) = 6$ $c = 1 + 2 = 3$ The product is: $abc = 6 \times 1 \times 3 = 18 \quad (\text{which is even}).$

2. Case 2: $b = 2$ $a = 8 - 2(2) = 4$ $c = 2 + 2 = 4$ The product is: $abc = 4 \times 2 \times 4 = 32 \quad (\text{which is even}).$

3. Case 3: $b = 3$ $a = 8 - 2(3) = 2$ $c = 3 + 2 = 5$ The product is: $abc = 2 \times 3 \times 5 = 30 \quad (\text{which is even}).$

The combinations can be summarized in the table below:

Since all possible combinations have been exhausted and in every case the product $abc$ is even, the product of $a$, $b$ and $c$ is always even.

我们可以通过代数结构分析变量的奇偶性： 由 $a = 8 - 2b = 2(4 - b)$ 可知，由于 $4-b$ 是整数，所以 $a$ 必定是 2 的倍数，即 $a$ 必定是一个偶数。 因为任意整数与偶数的乘积必然是偶数（$\text{even} \times \text{integer} = \text{even}$），所以只要 $a$ 是偶数，不管 $b$ 和 $c$ 的奇偶性如何，$abc$ 的乘积必然是偶数。

From the equations $c = b + 2$ and $a + b + c = 10$, we obtain:

Since $b$ is an integer, $4 - b$ is also an integer. Therefore, $a$ must be an even integer.

Since the product of any integers with at least one even factor is always even, the product $abc$ must be even for any valid integer combinations.

Hence, the product of $a$, $b$ and $c$ is always even.

将乘积 $abc$ 完全用单一变量 $b$ 表示出来：

提取公因子 2，写成 $2 \times \text{integer}$ 的形式，从而直接证明其整除性。

Using the equations $a = 8 - 2b$ and $c = b + 2$, we can express the product $abc$ in terms of $b$ only:

Since $b$ is an integer, $(4 - b)b(b + 2)$ is also an integer.

Since $abc$ has a factor of $2$, $abc$ is divisible by $2$, which means the product $abc$ is always even.