jmc

Multiplying the given equations together, we have $$\begin{array}{rl}(xyz{)}^2& =(-80-320i)\cdot 60\cdot (-96+24i)\\ & =-80(1+4i)\cdot 60\cdot -24(4-i)\\ & =(80\cdot 60\cdot 24)(8+15i).\end{array}$$To solve for $xyz,$ we find a complex number $a+bi$ whose square is $8+15i$ (where $a$ and $b$ are real); that is, we want $$(a+bi{)}^2=(a^2-b^2)+2abi=8+15i.$$Equating the real and imaginary parts, we get the equations $a^2-b^2=8$ and $2ab=15.$ Then $b=\frac{15}{2a};$ substituting into the other equation gives $a^2-\frac{225}{4a^2}=8,$ or $4a^4-32a^2-225=0.$ This factors as $$(2a^2-25)(2a^2+9)=0$$so $2a^2-25=0$ (since $a$ is real), and $a=\pm \frac{5}{\sqrt{2}}.$ Then $b=\frac{15}{2a}=\pm \frac{3}{\sqrt{2}}.$ Therefore, (arbitrarily) choosing both $a$ and $b$ positive, we have $$(xyz{)}^2=(80\cdot 60\cdot 24){\left(\frac{5}{\sqrt{2}}+\frac{3}{\sqrt{2}}i\right)}^2,$$and so $$\begin{array}{rl}xyz& =\pm \sqrt{80\cdot 60\cdot 24}\left(\frac{5}{\sqrt{2}}+\frac{3}{\sqrt{2}}i\right)\\ & =\pm 240\sqrt{2}\left(\frac{5}{\sqrt{2}}+\frac{3}{\sqrt{2}}i\right)\\ & =\pm 240(5+3i).\end{array}$$Then $$\begin{array}{rl}x& =\frac{xyz}{yz}=\pm \frac{240(5+3i)}{60}=\pm (20+12i),\\ z& =\frac{xyz}{xy}=\pm \frac{240(5+3i)}{-80(1+4i)}=\pm \frac{-3(5+3i)(1-4i)}{17}=\pm \frac{-3(17-17i)}{17}=\pm (-3+3i),\\ y& =\frac{xyz}{xz}=\pm \frac{240(5+3i)}{-24(4-i)}=\pm \frac{-10(5+3i)(4+i)}{17}=\pm \frac{-10(17+17i)}{17}=\pm (-10-10i).\end{array}$$Therefore, $x+y+z=\pm (7+5i),$ so $|x+y+z|=\sqrt{7^2+5^2}=\boxed{\sqrt{74}}.$