5. Given four points in space $A, B, C, D$ satisfying $A B \perp A C, A B \perp A D, A C \perp A D$, and $A B=A C=A D=1, Q$ is a moving point on the circumscribed sphere of the tetrahedron $A-B C D$, then the maximum distance from point $Q$ to the plane $B C D$ is
Answer $\frac{2 \sqrt{3}}{3}$.
Explanation Establish a spatial rectangular coordinate system with $A C, A D, A B$ as the $x, y, z$ axes, respectively, then $A(0,0,0), C(1,0,0), D(0,1,0), B(0,0,1)$. Let the center of the sphere be $O(x, y, z)$, and the radius be $R$, then $\left\{\begin{array}{l}x^{2}+y^{2}+z^{2}=R^{2} \\ (x-1)^{2}+y^{2}+z^{2}=R^{2} \\ x^{2}+(y-1)^{2}+z^{2}=R^{2} \\ x^{2}+y^{2}+(z-1)^{2}=R^{2}\end{array}\right.$, solving this gives $\left\{\begin{array}{l}x=\frac{1}{2} \\ y=\frac{1}{2} \\ z=\frac{1}{2} \\ R=\frac{\sqrt{3}}{2}\end{array}\right.$ Since $\triangle A C D$ is an equilateral triangle, the coordinates of its center are $O^{\prime}\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$, the maximum distance from point $Q$ to the plane $B C D$ is
$$
R+O O^{\prime}=\frac{\sqrt{3}}{2}+\sqrt{\left(\frac{1}{2}-\frac{1}{3}\right)^{2}+\left(\frac{1}{2}-\frac{1}{3}\right)^{2}+\left(\frac{1}{2}-\frac{1}{3}\right)^{2}}=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{6}=\frac{2 \sqrt{3}}{3}
$$

Therefore, the answer is $\frac{2 \sqrt{3}}{3}$.