Solve $$\begin{array}{l} 2 x-3 y+\frac{1}{x y}=6 \\ 3 z-6 x+\frac{1}{x z}=2 \\ 6 y-2 z+\frac{1}{y z}=3 \end{array}$$
Notice that $$-2(2x-3y)-\frac{2}{3}(3z-6x)=6y-2z$$
So we get: $$\frac{2}{xy}-12-\frac{4}{3}+\frac{2}{3xz}=3-\frac{1}{yz}$$ So we get $$\frac{2}{xy}+\frac{2}{3xz}+\frac{1}{yz}=\frac{49}{3}$$ $\implies$ $$\frac{3x+2y+6z}{xyz}=49$$ Any approach from here?