jmc

First, $x^2+4xy+4y^2=(x+2y{)}^2.$ By AM-GM, $$x+2y\ge 2\sqrt{2xy},$$so $(x+2y{)}^2\ge 8xy.$ Hence, $$x^2+4xy+4y^2+2z^2\ge 8xy+2z^2.$$If we apply AM-GM directly to $8xy$ and $2z^2,$ then ignoring constants, we will get the term $\sqrt{xy{z}^2}.$ But the condition is $xyz=32.$ So instead, we write $8xy+2z^2$ as $4xy+4xy+2z^2.$ Then by AM-GM, $$\begin{array}{rl}4xy+4xy+2z^2& \ge 3\sqrt[3]{(4xy)(4xy)(2z^2)}\\ & =3\sqrt[3]{32x^2{y}^2{z}^2}\\ & =3\sqrt[3]{32\cdot 32^2}\\ & =96.\end{array}$$Equality occurs when $x=2y$ and $4xy=2z^2.$ Along with the condition $xyz=32,$ we can solve to get $x=4,$ $y=2,$ and $z=4,$ so the minimum value is $\boxed{96}.$