\(3x^2-18y^2+2z^2+3y^2z^2-18x=27\Leftrightarrow3\left(x-3\right)^2+2z^2-18y^2+3y^2z^2=54\)(*)
Để phương trình có nghiệm nguyên thì \(z^2⋮3\Leftrightarrow z⋮3\Leftrightarrow z^2⋮9\Leftrightarrow z^2\ge9\)
Ta có (*)\(\Leftrightarrow3\left(x-3\right)^2+2z^2+3y^2\left(z^2-6\right)=54\Rightarrow54=3\left(x-3\right)^2+2z^2+3y^2\left(z^2-6\right)\ge3\left(x-3\right)^2+2.9+3y^2\Leftrightarrow3\left(x-3\right)^2+3y^2\le12\Leftrightarrow y^2\le4\Leftrightarrow y^2=1\) hoặc y2=4
_ y2=1\(\Leftrightarrow y=1\)
Vậy (*) có dạng \(3\left(x-3\right)^2+5z^2=72\Leftrightarrow5z^2\le72\Leftrightarrow z^2=9\Leftrightarrow z=3\Leftrightarrow x=6\)_y2=4\(\Leftrightarrow y=2\)
Vậy (*) có dạng \(3\left(x-3\right)^2+14z^2=126\Leftrightarrow14z^2\le126\Leftrightarrow z^2\le9\Leftrightarrow\)\(z=3\Leftrightarrow x=3\)
Vậy (x;y;z)={(3;2;3);(6;1;3)}