Let $C \subset \mathbb{P}^2$ be a planar conic curve, defined by a ternary quadratic form $Q(x\_1, x\_2, x\_3)$ say. Suppose that $C(\mathbb{Q}) \ne \emptyset$, or equivalently, that $C$ is everywhere locally soluble (i.e., $C(\mathbb{Q}\_p) \ne \emptyset$ for every prime $p$ and $C(\mathbb{R}) \ne \emptyset$). Further, it is easy to show that there exist primitive binary quadratic forms $f\_1, f\_2, f\_3 \in \mathbb{Z}[u,v]$ that parametrize the rational points $C(\mathbb{Q})$. In particular, $(x\_1, x\_2, x\_3) = (f\_1(u,v), f\_2(u,v), f\_3(u,v)), u,v \in \mathbb{Z}$.


For example, if $Q(x\_1, x\_2, x\_3) = x\_1^2 + x\_2^2 - x\_3^2$ then we can parametrize the primitive solutions by $x\_1 = 2uv, x\_2 = u^2 - v^2, x\_3 = u^2 + v^2$.


For a given ternary quadratic form $Q(x\_1, x\_2, x\_3) \in \mathbb{Z}[x\_1, x\_2, x\_3]$ which is everywhere locally soluble, call a triple of binary quadratic forms $(f\_1, f\_2, f\_3)$ a parametrizing triple if $Q(f\_1, f\_2, f\_3) \equiv 0$. Consider the binary sextic form $F\_Q = f\_1 f\_2 f\_3$.


In general, what can we say about $F\_Q$, given $F$? In the example above, we have $(f\_1, f\_2, f\_3) = (2uv, u^2 - v^2, u^2 + v^2)$ and $F\_Q = 2uv(u^2 - v^2)(u^2 + v^2)$. This $F\_Q$ is very special: indeed, it is a sextic *Klein form*, with an exceptionally large $\text{PGL}\_2$ automorphism group. Does this hold in general when $Q$ is a diagonal form?



Conics and triples of binary quadratic forms
I don't know if the following will fully answer your question, since I'm not sure if you attach a huge importance to have integral coefficients/primitive solutions. Apologies if it doesn't.


Let $Q$ be **any** nondegenerate (I assume this is the case of interest) ternary quadratic form such that $Q(\mathbb{Q})\neq 0$.
Then the theory of quadratic forms says that $Q$ splits off a hyperbolic plane. Hence $Q\simeq \langle a,1,-1\rangle$ for some nonzero $a$.


Now it is known that $\langle 1,-1\rangle\simeq \langle a,-a\rangle$ for any nonzero $a$, so finally $Q$ is isomorphic to $\langle a ,a ,-a\rangle$.


In other words, there exists a basis of $K^3$ such that $Q(x\_1,x\_2,x\_3)=a(x\_1^2+x\_2^2-x\_3^2)$, where the $x\_i'$s are the coordinates of a vector $x\in K^3$ in this basis.


Now $Q(x\_1,x\_2,x\_3)=0\iff x\_1^2+x\_2^2-x\_3^2=0$.


All in all, the case you already covered is the only one. Hope this helps.


