Given the spatial points $A, B, C, D, E, F$, we know that $A C=A D=B E=B F=10$, $A E=A F=B C=B D=12$, $C D=E F=11$, $C E=D F=5$. What can the distance $A B$ be? 
Hang the $CD$ edge vertically in front of us so that the vertex $A$ of the body is to the left of it, and the vertex $B$ is to the right. The triangles $ACD$ and $BCD$ are isosceles, with heights denoted by $a$ and $b$ respectively, the sought $AB$ distance is denoted by $c$, and the midpoint of $CD$ is denoted by $G$. In the triangle $ABG$, in every case, it holds that

$$
b-a<8.8
$$

This is sharper than the lower bound in (1), since $a=\sqrt{69.75}=8.35, b=\sqrt{113.75}=10.67$, so we finally get that under the given conditions

$$
8.8<AB<\sqrt{69.75}+\sqrt{113.75}=19.2
$$

and conversely, if (3) holds for $AB$, then there exist points $C, D, E, F$ such that the given distances occur between them.

Remark. Our problem is related to the article by László Csirmaz titled "Can someone move if we dressed them in a suit made of planes?" (KöMaL, November 1979, Volume 59, No. 3-4, p. 97). A typo crept into the first publication: instead of the segment $EF$, we stated that the segment $ED$ was 11 units long. In that case, only two values, 9.12 and 18.94, could have been valid for $AB$ (KöMaL, April 1980, Volume 60, No. 4, pp. 155-158).