jmc

We can apply Simon's Favorite Factoring Trick to each of the equations. Indeed, re-arranging, $$\begin{array}{rl}xy+2x+3y& =6,\\ yz+4y+2z& =6,\\ xz+4x+3z& =30,\end{array}$$Adding $6$, $8$, and $12$ to both sides of each equation, respectively, yields $$\begin{array}{rl}xy+2x+3y+6=(x+3)(y+2)& =12,\\ yz+4y+2z+8=(y+2)(z+4)& =14,\\ xz+4x+3z+12=(x+3)(z+4)& =42,\end{array}$$At this point, we can substitute and solve by elimination. Even simpler, notice that if we take the product of all three equations, we obtain $$[(x+3)(y+2)(z+4){]}^2=12\cdot 14\cdot 42=2^4\cdot 3^2\cdot 7^2,$$so $$(x+3)(y+2)(z+4)=\pm 2^2\cdot 3\cdot 7.$$We can now substitute that $(y+2)(z+4)=14$ to find that $$(x+3)(y+2)(z+4)=14(x+3)=\pm 2^2\cdot 3\cdot 7.$$Hence, $x+3=\pm 6,$ so $x$ is $3$ or $-9.$ The positive root is thus $x=\boxed{3}$.