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By Cauchy-Schwarz, $$(y+z-2)+(z+x-4)+(x+y-6)]\left[\frac{(x+2{)}^2}{y+z-2}+\frac{(y+4{)}^2}{z+x-4}+\frac{(z+6{)}^2}{x+y-6}\right]\ge [(x+2)+(y+4)+(z+6){]}^2.$$This simplifies to $$36(2x+2y+2z-12)\ge (x+y+z+12{)}^2.$$Let $s=x+y+z.$ Then $36(2s-12)\ge (s+12{)}^2.$ This simplifies to $s^2-48s+576\le 0,$ which then factors as $(s-24{)}^2\le 0.$ Hence, $s=24.$

Thus, the inequality above turns into an equality, which means $$\frac{x+2}{y+z-2}=\frac{y+4}{z+x-4}=\frac{z+6}{x+y-6}.$$Since $x+y+z=24,$ $$\frac{x+2}{22-x}=\frac{y+4}{20-y}=\frac{z+6}{18-z}.$$Each fraction must then be equal to $$\frac{(x+2)+(y+4)+(z+6)}{(22-x)+(20-y)+(18-z)}=\frac{x+y+z+12}{60-(x+y+z)}=1.$$From here, it is easy to solve for $x,$ $y,$ and $z,$ to find $x=10,$ $y=8,$ and $z=6.$

Hence, $(x,y,z)=\boxed{(10,8,6)}.$