Comparing These

Clearly $B \subseteq A$ because the coordinates of elements of $B$ all have the same parity. To show that $A \subseteq B,$ consider $(a,b,c) \in A$ and let $\begin{aligned} \alpha &= \frac{b+c}2 \\ \beta &= \frac{a-b}2 \\ \gamma &= \frac{a-c}2 \end{aligned}$ Then $\alpha, \beta, \gamma$ are all integers because $b+c, a-b, a-c$ are all even, and it is easy to check that $(a,b,c) = (\alpha + \beta + \gamma, \alpha - \beta + \gamma, \alpha + \beta - \gamma).$ So $A = B.$

I will show equality by showing inclusion in both directions. I show any arbitrary element of $B$ is also in $A$ , and vice versa.

An arbitrary element of B can be expressed in the form above, with all three elements of the ordered triple expressed as linear sums of three arbitrary integers, $\alpha, \beta,$ and $\gamma$ .

The $B \rightarrow A$ inclusion is clear enough:

$a = \alpha + \beta + \gamma$ ,

$b = \alpha - \beta + \gamma$ ,

$c = \alpha + \beta - \gamma$

Do we know that all three of these are of the same parity? Yes, because $(\beta + \gamma)$ , $(\beta - \gamma)$ , and $(-\beta + \gamma)$ are all of the same parity. Adding $\alpha$ to those three numbers will result in three sums all of the same parity. Thus $B \subset A$ .

To show $A \subset B$ , we given a triple $(a,b,c) \in A$ , we need corresponding values of $\alpha, \beta,$ and $\gamma$ . The $A \rightarrow B$ mapping is also simple, though it isn't explicitly spelled out.

$\alpha = \frac{b+c}{2}$

$\beta = \frac{a-b}{2}$

$\gamma = \frac{a -c}{2}$

Since $a, b$ , and $c$ are all of the same parity, $(b+c), (a-b)$ and $(a-c)$ are all even numbers, and thus $\alpha, \beta$ and $\gamma$ are all integers. Thus this mapping shows $(a,b,c) \in B$ .

So $A \subset B$ and $B \subset A$ , i.e., $A = B$ .

For any natural $α,β,\gamma$ ; $α+β+\gamma, α-β+\gamma, α+β-\gamma$ are of the same parity. So $\boxed {A=B}$