Find $xyz$ if $x^2+2y^2+2z^2-2x-6y-10z+2xy+2yz+14=0$

Question

The question is

Find the sum of all possible values of $xyz$ given that $x, y, z\in \Bbb Z$ satisfying $$x^2+2y^2+2z^2-2x-6y-10z+2xy+2yz+14=0.$$

Some thought so far:

Obviously $x$ is even. I assume $x=2k$, but I don't know how to proceed. I tried to complete squares and it became more complicated: $$(x+y-1)^2+(y+z+4)^2+(z-9)^2=82.$$

WolframAlpha gives $48$ set of solutions, so I think there are ways calculating sum of all possible values of $xyz$ without finding all solutions.

Answer 1

Completing the squares: $$(x+y-1)^2+(y+z-2)^2+(z-3)^2=0 \Rightarrow \\ \begin{cases}x+y-1=0\\ y+z-2=0\\ z-3=0\end{cases} \Rightarrow \\ (x,y,z)=(2,-1,3) \Rightarrow xyz=-6.$$

Answer 2

Completing the squares was great idea! Now write $82$ as the sum of three integer squares. There aren't many ways to do that. Each way gives you a linear system for $x,y,z$, which may or may not have integer solutions.