<s>[INST] Find $xyz$ if $x^2+2y^2+2z^2-2x-6y-10z+2xy+2yz+14=0$ 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.
 [/INST] Completing the squares:
$$(x+y-1)^2+(y+z-2)^2+(z-3)^2=0 \Rightarrow \\
\begin{cases}x+y-1=0\\ y+z-2=0\\ z-3=0\end{cases} \Rightarrow \\
(x,y,z)=(2,-1,3) \Rightarrow xyz=-6.$$
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