jmc

Let $a=(x-y)/3,$ $b=(y-z)/2,$ and $c=z.$ Then $x-y=3a,$ $y-z=2b,$ and $z=c.$ Adding these, we get $x=3a+2b+c.$ Hence, $$x+\frac{108}{(x-y{)}^3(y-z{)}^{2}z}=3a+2b+c+\frac{1}{a^3{b}^{2}c}.$$By AM-GM, $$a+a+a+b+b+c+\frac{1}{a^3{b}^{2}c}\ge 7.$$Equality occurs when $a=b=c=1,$ or $x=6,$ $y=3,$ and $z=1,$ so the minimum value is $\boxed{7}.$