The sum $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5 + \alpha_6$ is even, if and only if, the number of $1$'s is even, that is $0$, $2$, $4$, $6$.

In the case of zero $1$'s, the possible selections are $2^6$, because for each $\alpha_i$ we have $2$ selections, ($0$ or $2$).

When we have two $1$'s, then they can be selected in $\binom{6}{2}$ ways and the rest four places can be completed by $2^4$ ways. It means that we have $2^4 \cdot \binom{6}{2}$ possible $6$-tuples.

Similarly, in the case we have four $1$'s we conclude that we have $2^2 \cdot \binom{6}{4}$ possible $6$-tuples.

When we have six $1$'s then obviously we have only one possible $6$-tuple.

Therefore the possible $6$-tuples are totally:
$$2^6 + 2^4 \cdot \binom{6}{2} + 2^2 \cdot \binom{6}{4} + 1 = 64 + 16 \cdot 15 + 4 \cdot 15 + 1 = 365.$$