Let $C$ be the conic in $\mathbb{P}^2$ given by $ax^2 + by^2 + cz^2 = 0$ with $a,b,c\in\mathbb{Q}$.
(every genus 0 curve over $\mathbb{Q}$ can be given this way right?)
Suppose $C$ has no rational points. Could there exist a solution $(\alpha,\beta,1)$ with $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = 3$?
Let me begin by addressing your parenthetical question.
Let $C/\mathbb{Q}$ be a genus $0$ smooth projective geometrically integral curve. Then, we know that $\omega:=\omega_{C/\mathbb{Q}}$ by Riemann-Roch has degree $2(0)-2=-2$ and thus $\omega^{-1}$ has degree $2$. Since $2\geqslant 2(0)+1$ we know that $\omega$ is very ample and by Riemann-Roch
$$h^0(\omega^{-1})-h^0(\omega^2)=2+1$$
so $h^0(\omega)=3$. Thus, we obtain a closed embedding of $C$ into $\mathbb{P}^2$. This is a hypersurface and so of the form $V(f)$ for some homogenous form $f$. It must be that $\deg f=2$ or $\deg f=1$ since $$0=\frac{(\deg f-1)(\deg f - 2)}{2}$$ by elementary considerations.
Thus, $C$ is isomorphic to $\mathbb{P}^1$ or a quadratic hypersurface in $\mathbb{P}^2$—the latter obviously being the case of $C(\mathbb{Q})=0$.
Now, as for Mohan's observation, if $C$ has a degree $3$ point $P$ then $\mathcal{O}(P)\otimes\omega$ has degree $1$. But, again, since $2(0)+1\geqslant 1$ this would force $\mathcal{O}(P)\otimes\omega$ to be very ample and since $h^0(\mathcal{O}(P)\otimes\omega)=2$ (by Riemann-Roch) this would embed $C$ into $\mathbb{P}^1$ or, in other words, $C$ would be isomorphic to $\mathbb{P}^1$ which is impossible since $C(\mathbb{Q})=\varnothing$.
So, to summarize. Every genus $0$ curve over $\mathbb{Q}$ (or any field) has a point of degree $2$ which embeds it as a quadratic/linear hypersurface in $\mathbb{P}^2$. It can be embedded as a linear hypersurface if and only if it has a degree $1$ point in which case it's $\mathbb{P}^1$.
This is of course related to $\mathrm{Br}(\mathbb{Q})$ since $C$ gives us a non-trivial class (a quaternion algebra in fact) if $C(\mathbb{Q})=\varnothing$
