66. Two equal triangles $K L M$ and $K L N$ have a common side $K L, \quad \widehat{K L M}=\widehat{L K N}=\pi / 3, \quad|K L|=a$, $|L M|=|K N|=6 a$. The planes $K L M$ and $K L N$ are mutually perpendicular. A sphere touches the segments $L M$ and $K N$ at their midpoints. Find the radius of the sphere.
66. Let $O_{1}$ and $O_{2}$ be the projections of the center of the sphere $O$ onto the planes $K L M$ and $K L N$, and let $P$ be the midpoint of $M L$.

The projections of $O_{1}$ and $O_{2}$ onto $K L$ should coincide. It can be proven that these projections fall on the midpoint of $K L$ - point $Q$ (Fig. 4). Since the dihedral angle between the planes $K L M$ and $K L N$ is $90^{\circ}$, the radius of the desired sphere will be

$$
\sqrt{\left.P P O_{1}\right|^{2}+\left|O_{1} Q\right|^{2}} .
$$

If we extend $O_{1} P$ until it intersects the line $K L$ at point $R$, then from the right triangle $P L R$ we find $|R L|=6 a$,
$|R P|=3 a \sqrt{3}$. Then we find

$$
\begin{aligned}
& |R Q|=\frac{11 a}{2}, \quad\left|O_{1} Q\right|=\frac{11 a \sqrt{3}}{6}-\left|R O_{1}\right|=\frac{11 a \sqrt{3}}{3}, \\
& \left|P O_{1}\right|=\frac{11 a \sqrt{3}}{3}-3 a \sqrt{3}=\frac{2 a \sqrt{\overline{3}}}{3} .
\end{aligned}
$$

Therefore, the radius of the sphere is

$$
\sqrt{\frac{4 a^{2}}{3}+\frac{121 a^{2}}{12}}=\frac{a}{2} \sqrt{\frac{137}{3}}
$$