jmc

By AM-GM, $$a+b\ge 2\sqrt{ab},$$so $(a+b{)}^2\ge 4ab.$

Also by AM-GM, $$(a+2c)+(b+2c)\ge 2\sqrt{(a+2c)(b+2c)},$$so $(a+b+4c{)}^2\ge 4(a+2c)(b+2c).$

Hence, $$\begin{array}{rl}(a+b{)}^2+(a+b+4c{)}^2& \ge 4ab+4(a+2c)(b+2c)\\ & =8ab+8ac+8bc+16c^2\\ & =8(ab+ac+bc+2c^2).\end{array}$$By AM-GM, $$\begin{array}{rl}ab+ac+bc+2c^2& =\frac{ab}{2}+\frac{ab}{2}+ac+bc+2c^2\\ & \ge 5\sqrt[5]{\frac{ab}{2}\cdot \frac{ab}{2}\cdot ac\cdot bc\cdot 2c^2}\\ & =5\sqrt[5]{\frac{a^3{b}^3{c}^4}{2}}.\end{array}$$Also by AM-GM, $$\begin{array}{rl}a+b+c& =\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c\\ & \ge 5\sqrt[5]{\frac{a}{2}\cdot \frac{a}{2}\cdot \frac{b}{2}\cdot \frac{b}{2}\cdot c}\\ & =5\sqrt[5]{\frac{a^2{b}^{2}c}{16}}.\end{array}$$Hence, $$\begin{array}{rl}\frac{(a+b+c)[(a+b{)}^2+(a+b+4c{)}^2]}{abc}& \ge 8\cdot \frac{5\sqrt[5]{\frac{a^2{b}^{2}c}{16}}\cdot 5\sqrt[5]{\frac{a^3{b}^3{c}^4}{2}}}{abc}\\ & =100.\end{array}$$Equality occurs when $a=b=2$ and $c=1,$ so the minimum value is $\boxed{100}.$