First of all, it is easy to see that:
$$\alpha_1\sqrt{2} + \alpha_2\sqrt{3} = \alpha_3\sqrt{2} + \alpha_4\sqrt{3}, \text{ with } \alpha_1, \alpha_2, \alpha_3, \alpha_4 \in \mathbb{Q}^* \Leftrightarrow \alpha_1 = \alpha_3 \text{ and } \alpha_2 = \alpha_4.$$
In fact, we can write the first relation in the form
$$(\alpha_1 - \alpha_3)\sqrt{2} = (\alpha_4 - \alpha_2)\sqrt{3},$$
and if $\alpha_1 - \alpha_3 \neq 0$, then $\frac{\alpha_4 - \alpha_2}{\alpha_1 - \alpha_3} = \frac{\sqrt{3}}{\sqrt{2}}$, absurd. Hence $\alpha_1 = \alpha_3$ and $\alpha_2 = \alpha_4$.
The converse is clear.
Let now $\frac{\alpha\sqrt{2} + \beta\sqrt{3}}{\beta\sqrt{2} + \gamma\sqrt{3}} = \kappa \in \mathbb{Q}$. Then $\alpha = \kappa\beta$ and $\beta = \kappa\gamma$, that is
$$\frac{\alpha}{\beta} = \frac{\beta}{\gamma} \Leftrightarrow \beta^2 = \alpha\gamma.$$
Hence we have:

$$\begin{aligned} \alpha^2 + \beta^2 + \gamma^2 &= \alpha^2 + \alpha\gamma + \gamma^2 = \alpha^2 + 2\alpha\gamma + \gamma^2 - \alpha\gamma \\ &= (\alpha + \gamma)^2 - \beta^2 = (\alpha + \beta + \gamma)(\alpha - \beta + \gamma), \end{aligned}$$
and so $B = \alpha - \beta + \gamma$.