I am interested in the following question. Let $q$ be a prime power and let $\mathbb{F}\_q$ be the finite field of cardinality $q$. Suppose $q>61$. Is it true that, for every $b\in \mathbb{F}\_q$ and for every $c\in \mathbb{F}\_q$ with $c\ne 0$, there exists $x,y,z\in\mathbb{F}\_q$ such that
\begin{align\*}
x^2+y^2+z^2&=b\\
xyz&=c.
\end{align\*}
I have computational evidence towards this and in fact the request $q>61$ is suggested by the computational data.



On a certain equation in finite fields
Generically the intersection of the surfaces described by these two equations is an algebraic curve of genus $4$. Once one has made sure that this curve is absolutely irreducible, by Weil there are $\mathbb F\_q$-points provided that $q$ is big enough. Weil assumes a smooth curve, so it can be a pain to handle singularities. However, there are explicit bounds, like Theorem 5.4.1 in the third edition of Fried and Jarden's [Field Arithmetic](https://doi.org/10.1007/b138352): If the absolutely irreducible affine curve has degree $d$, then the number of $\mathbb F\_q$-points is at least $q+1-(d-1)(d-2)\sqrt{q}-d$. In your case $d=6$. This lower bound is positive once $q\ge419$. So one has to check the smaller cases directly, or has to resort to Joe Silverman's suggestion from the comments.


