Multiply $v_1$ to both sides above, and take advantage of orthogonality & normality.
$$A v_1 = \alpha v_1 (v_1^{T}v_1) + \beta v_2 (v_2^{T}v_1)$$
$$ A v_1 = \alpha v_1 (1) + \beta v_2 (0)$$
$$ A v_1 = \alpha v_1 $$
Do the same for $v_2$. You'll find $\alpha$ & $\beta$ are eigenvalues. So, A is correct.