jmc

Hoping for cancellation, we first compute $\frac{1}{a}+\frac{1}{d},$ since $a$ and $d$ have two opposite signs: $$\begin{array}{rl}\frac{1}{a}+\frac{1}{d}& =\frac{a+d}{ad}\\ & =\frac{(\sqrt{2}+\sqrt{3}+\sqrt{6})+(-\sqrt{2}-\sqrt{3}+\sqrt{6})}{(\sqrt{2}+\sqrt{3}+\sqrt{6})(-\sqrt{2}-\sqrt{3}+\sqrt{6})}\\ & =\frac{2\sqrt{6}}{(\sqrt{6}{)}^2-(\sqrt{2}+\sqrt{3}{)}^2}\\ & =\frac{2\sqrt{6}}{1-2\sqrt{6}}.\end{array}$$Similar cancellation occurs when adding $\frac{1}{b}+\frac{1}{c}$: $$\begin{array}{rl}\frac{1}{b}+\frac{1}{c}& =\frac{b+c}{bc}\\ & =\frac{(-\sqrt{2}+\sqrt{3}+\sqrt{6})+(\sqrt{2}-\sqrt{3}+\sqrt{6})}{(-\sqrt{2}+\sqrt{3}+\sqrt{6})(\sqrt{2}-\sqrt{3}+\sqrt{6})}\\ & =\frac{2\sqrt{6}}{(\sqrt{6}{)}^2-(\sqrt{2}-\sqrt{3}{)}^2}\\ & =\frac{2\sqrt{6}}{1+2\sqrt{6}}.\end{array}$$It follows that $$\begin{array}{rl}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}& =\frac{2\sqrt{6}}{1-2\sqrt{6}}+\frac{2\sqrt{6}}{1+2\sqrt{6}}\\ & =\frac{4\sqrt{6}}{1^2-(2\sqrt{6}{)}^2}\\ & =-\frac{4\sqrt{6}}{23},\end{array}$$so ${\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}^2=\boxed{\frac{96}{529}}.$