Giải:
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow\left\{\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)
Ta có: \(a^2+3b^2-2c^2=\left(-16\right)\)
\(\Rightarrow4k^2+27k^2-32k^2=-16\)
\(\Rightarrow\left(-1\right)k^2=-16\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=\pm4\)
+) \(k=4\Rightarrow a=8;b=12;c=16\)
+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)