jmc

We write $$\begin{array}{rl}\frac{(1+5z)(4z+3x)(5x+6y)(y+18)}{xyz}& =\frac{4}{5}\cdot \frac{(1+5z)(5z+\frac{15}{4}x)(5x+6y)(y+18)}{xyz}\\ & =\frac{4}{5}\cdot \frac{4}{3}\cdot \frac{(1+5z)(5z+\frac{15}{4}x)(\frac{15}{4}z+\frac{9}{2}y)(y+18)}{xyz}\\ & =\frac{4}{5}\cdot \frac{4}{3}\cdot \frac{2}{9}\cdot \frac{(1+5z)(5z+\frac{15}{4}x)(\frac{15}{4}x+\frac{9}{2}y)(\frac{9}{2}y+81)}{xyz}\\ & =\frac{32}{135}\cdot \frac{(1+5z)(5z+\frac{15}{4}x)(\frac{15}{4}x+\frac{9}{2}y)(\frac{9}{2}y+81)}{xyz}.\end{array}$$Let $a=5z,$ $b=\frac{15}{4}x,$ and $c=\frac{9}{2}y,$ so $z=\frac{1}{5}a,$ $x=\frac{4}{15}b,$ and $y=\frac{2}{9}c.$ Then $$\begin{array}{rl}\frac{32}{135}\cdot \frac{(1+5z)(5z+\frac{15}{4}x)(\frac{15}{4}x+\frac{9}{2}y)(\frac{9}{2}y+81)}{xyz}& =\frac{32}{135}\cdot \frac{(1+a)(a+b)(b+c)(c+81)}{\frac{4}{15}b\cdot \frac{2}{9}c\cdot \frac{1}{5}a}\\ & =20\cdot \frac{(1+a)(a+b)(b+c)(c+81)}{abc}\\ & =20\cdot (1+a)\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{81}{c}\right).\end{array}$$By AM-GM, $$\begin{array}{rl}1+a& =1+\frac{a}{3}+\frac{a}{3}+\frac{a}{3}\ge 4\sqrt[4]{{\left(\frac{a}{3}\right)}^3},\\ 1+\frac{b}{a}& =1+\frac{b}{3a}+\frac{b}{3a}+\frac{b}{3a}\ge 4\sqrt[4]{{\left(\frac{b}{3a}\right)}^3},\\ 1+\frac{c}{b}& =1+\frac{c}{3b}+\frac{c}{3b}+\frac{c}{3b}\ge 4\sqrt[4]{{\left(\frac{c}{3b}\right)}^3},\\ 1+\frac{81}{c}& =1+\frac{27}{c}+\frac{27}{c}+\frac{27}{c}\ge 4\sqrt[4]{{\left(\frac{27}{c}\right)}^3},\end{array}$$so $$\begin{array}{rl}20\cdot (1+a)\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{81}{c}\right)& \ge 20\cdot 256\sqrt[4]{{\left(\frac{a}{3}\right)}^3\cdot {\left(\frac{b}{3a}\right)}^3\cdot {\left(\frac{c}{3b}\right)}^3\cdot {\left(\frac{27}{c}\right)}^3}\\ & =5120.\end{array}$$Equality occurs when $$1=\frac{a}{3}=\frac{b}{3a}=\frac{c}{3b}=\frac{27}{c},$$or $a=3,$ $b=9,$ and $c=27,$ which means $x=\frac{12}{5},$ $y=6,$ and $z=\frac{3}{5}.$ Therefore, the minimum value is $\boxed{5120}.$