Seven, in plane $\alpha$ there is an equilateral $\triangle \mathrm{A}^{\prime} \mathrm{BC}, \mathrm{DE} / / \mathrm{BC}$, fold $\triangle \mathrm{A}^{\prime} \mathrm{DE}$ to the position of $\triangle \mathrm{ADE}$, making plane $\mathrm{ADE}$ perpendicular to plane $\alpha$, ask where DE should be taken so that the distance $\mathrm{AB}$ is the shortest.
Seven, Solution: Let $\mathrm{A}^{\prime} \mathrm{M}$ be the altitude of $\triangle \mathrm{ABC}$ $\Rightarrow \mathrm{AN} \perp \mathrm{DE}$.
$\because$ Plane $\mathrm{ADE}$ is on plane $\alpha, \therefore \mathrm{AN}$ is perpendicular to plane $\alpha$.
Let the side length of the equilateral $\triangle \mathrm{A}^{\prime} \mathrm{BC}$ be
$\mathrm{a}$, then $\mathrm{A}^{\prime} \mathrm{M}=\sqrt{3} \mathrm{a}$.
Let $\mathrm{AN}=x$.
$\angle$
$M N=\frac{\sqrt{2}}{2} a-x$,
By $\mathrm{AB}^{2}=\mathrm{AN}^{2}+\mathrm{MN}^{2}+\mathrm{MB}^{2}$
$$
\begin{array}{l}
=x^{2}+\left(\frac{\sqrt{3}}{2} a-x\right)^{2}+\left(\frac{a}{2}\right)^{2} \\
=2\left(x-\frac{\sqrt{3}}{4} a\right)^{2}+\frac{5}{8} a^{2},
\end{array}
$$

Therefore, when $\mathrm{x}=\frac{\sqrt{3}}{4} \mathrm{a}$, the distance $\mathrm{AB}$ is the shortest, with a value of $\sqrt{\frac{5}{8}} \mathrm{a}$, i.e., when $D E$ passes through the midpoint of the height, the distance $\mathrm{AB}$ is the shortest.