Finding every solution of $a^2+b^2+c^2=3$ in $\mathbb{Q}(i)$

Question

Specifically, $a,b,c\in\mathbb{Q}(i)$ are complex numbers with rational parts whose squares sum to $a^2+b^2+c^2=3$. There's an answer to this question over $\mathbb{Q}$ already here but I couldn't find any more information.

Like, does the parametrization over $\mathbb{Q}$ still give you all solutions over $\mathbb{Q}(i)$ if you let the parameters vary over $\mathbb{Z}[i]$ instead of $\mathbb{Z}$?

Here's the substitutions for the parametrization from the linked post: $$a=\frac{r^2+s^2-t^2-2t(r+s)}{r^2+s^2+t^2}\quad b=\frac{r^2-s^2+t^2-2s(r+t)}{r^2+s^2+t^2}\quad c=\frac{-r^2+s^2+t^2-2r(s+t)}{r^2+s^2+t^2}$$ $$r=\frac{1}{b-1}\quad s=\frac{1}{c-1}\quad t=\frac{a-1}{(b-1)(c-1)}$$

My next bet is to try to find research papers on complex ternary quadratic forms. Either that or to try to do something with "pure" quaternions which I don't really understand.

I also spent some time studying the proof for the parametrization of Pythagorean triples over both $\mathbb{Z}$ and $\mathbb{Z}[i]$ and feel pretty comfortable with it. But I couldn't find anything there that was adaptable to this diophantine problem.

I did notice that you can scale the equation into $a^2+b^2+c^2=3d^2$ over $\mathbb{Z}[i]$ instead of $\mathbb{Q}(i)$ by clearing denominators. And that this rearranges to $$(a+bi)(a-bi)=(\sqrt{3}d+c)(\sqrt{3}d-c)$$ which gives a factorization of one number in two ways and that seems useful. But this would be a factorization in $\mathbb{Z}[i,\sqrt{3}]$ which might be useless as unique factorization might be messed up. Some other readings made it seem like $\mathbb{Z}[i,\sqrt{3}]$ is actually just the cyclotomic ring $\mathbb{Z}[\zeta_{12}]$ in disguise so I thought to spend some effort on that connection eventually.

Any guidance or help is appreciated, thank you.