This is a sequel of this question where I asked for which positive integer $n$ the
set of primes of the former $x^2+ny^2$ was defined by congruences (a set of primes $P$ is *defined by congruences* if there is a positive integer $d$ and a subset $A$ of $\mathbb{Z}/d\mathbb{Z}$ such that a prime $p$ is in $P$ if and only if $p$ mod $d$ is in $A$, up to a finite number of exceptions). I was taught there that the answer was "exactly when $n$ is idoneal", that there is finitely many idoneal numbers, and that all are known but perhaps one.

When is the set of primes of the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ defined by congruences?

My motivation is not just an idle ternary generalization of the binary case. I really met this question while working on a problem concerning modular forms, and also the slightly more general question, given a fixed positive integer $a$: when is the set of primes $p$ such that $ap$ has the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ defined by congruences?

I am well aware that since the set of integers represented by a ternary quadratic form is not stable by multiplication, it is much less natural to ask the question for prime numbers instead of all positive integers than in the case of a binary quadratic form. Yet this is really the question for primes that appears in my study (for about a dozen specific ternary forms, actually).

I have found a very interesting paper by Dickson (Ternary quadratic forms and congruences. Ann. of Math. (2) 28 (1926/27), no. 1-4, 333–341.) which solves the question for the integers represented by $x^2+ny^2+mz^2$: there is only a finite explicit numbers of $(n,m)$ such that this set of integers is defined by congruences (in the obvious sense). But the proof does not seem (to me) to be easily generalizable to primes. Other mathscinet research did not give me any more informations.

When I try to think to the question, I meet an even more basic (if perhaps slightli more sophisticated) question that I can't answer:

When is the set of primes of the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ Frobenian? (Is it "always"?)

A set of primes $P$ is called *Frobenian* (a terminology probably introduced by Serre) if there is a finite Galois extension $K/\mathbb{Q}$, and a subset $A$ of Gal$(K/\mathbb{Q})$ stable by conjugacy such that a prime $p$ is in $P$ if and only if Frob${}_p \in A$, except for a finite number of exceptions. A set determined by congruences is a Frobenian set for which
we can take $K$ cyclotomic over $\mathbb{Q}$, which is the same by Kronecker-Weber as abelian over $\mathbb{Q}$. For a quadratic binary
quadratic form (for example $x^2+ny^2$), the set of represented primes is always Frobenian ($K$ can be taken as the
ring class field of $\mathbb{Z}[\sqrt{-n}]$, and $A=\{1\}$, as explained in Cox's book). But I fail to see the reason
(which may nevertheless be trivial) for which the same result would be true for a general ternary quadratic form.
I should add that for my specific ternary forms, I can show that the set is Frobenian, but I am not sure how to extend the argument to all ternary quadratic forms.

Finally, let me say that I would be interested in any book, survey or references on this kind of question (which surely must have been studied), and that I am also interested in analog questions for quaternary quadratic forms (which might be easier, because of multiplicative properties related to quaternions).