10,11

All pairwise distances between four points in space are equal to 1. Find the distance from one of these points to the plane defined by the other three.
Let $A, B, C$ and $D$ be points, the pairwise distances between which are equal to 1. Consider the triangular pyramid $A B C D$ with vertex $D$. Its base is an equilateral triangle $A B C$. The lateral edges $D A, D B$ and $D C$ of this pyramid are equal to each other, so its height $D O$ passes through the center $O$ of the circumcircle of the base $A B C$, i.e., through the center of the equilateral triangle $A B C$ with side 1. Let $M$ be the midpoint of side $B C$. Then

$$
A M=A B \sin \angle A B M=A B \sin 60^{\circ}=1 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}
$$

$$
A O=\frac{2}{3} A M=\frac{2}{3} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}
$$

Since $D O$ is the height of the pyramid, the distance from point $D$ to the plane $A B C$ is equal to the length of segment $D O$. From the right triangle $A O D$, we find that

$$
D O=\sqrt{A D^{2}-A O^{2}}=\sqrt{1-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}
$$

It is clear that the other desired distances are also equal to $\sqrt{\frac{2}{3}}$.

Answer

$\sqrt{\frac{2}{3}}$