jmc

Our strategy is to take $x^2+y^2+z^2$ and divide into several expression, apply AM-GM to each expression, and come up with a multiple of $2xy\sqrt{6}+8yz.$

Since we want terms of $xy$ and $yz$ after applying AM-GM, we divide $x^2+y^2+z^2$ into $$(x^2+k{y}^2)+[(1-k)y^2+z^2].$$By AM-GM, $$\begin{array}{rl}{x}^2+k{y}^2& \ge 2\sqrt{(x^2)(k{y}^2)}=2xy\sqrt{k},\\ (1-k)y^2+z^2& \ge 2\sqrt{((1-k)y^2)(z^2)}=2yz\sqrt{1-k}.\end{array}$$To get a multiple of $2xy\sqrt{6}+8yz,$ we want $k$ so that $$\frac{2\sqrt{k}}{2\sqrt{6}}=\frac{2\sqrt{1-k}}{8}.$$Then $$\frac{\sqrt{k}}{\sqrt{6}}=\frac{\sqrt{1-k}}{4}.$$Squaring both sides, we get $$\frac{k}{6}=\frac{1-k}{16}.$$Solving for $k,$ we find $k=\frac{3}{11}.$

Thus, $$\begin{array}{rl}{x}^2+\frac{3}{11}{y}^2& \ge 2xy\sqrt{\frac{3}{11}},\\ \frac{8}{11}{y}^2+z^2& \ge 2yz\sqrt{\frac{8}{11}}=4yz\sqrt{\frac{2}{11}},\end{array}$$so $$1=x^2+y^2+z^2\ge 2xy\sqrt{\frac{3}{11}}+4yz\sqrt{\frac{2}{11}}.$$Multiplying by $\sqrt{11},$ we get $$2xy\sqrt{3}+4yz\sqrt{2}\le \sqrt{11}.$$Multiplying by $\sqrt{2},$ we get $$2xy\sqrt{6}+8yz\le \sqrt{22}.$$Equality occurs when $x=y\sqrt{\frac{3}{11}}$ and $y\sqrt{\frac{8}{11}}=z.$ Using the condition $x^2+y^2+z^2=1,$ we can solve to get $x=\sqrt{\frac{3}{22}},$ $y=\sqrt{\frac{11}{22}},$ and $z=\sqrt{\frac{8}{22}}.$ Therefore, the maximum value is $\boxed{\sqrt{22}}.$