109. A sphere of radius 2 is inscribed in a regular triangular pyramid $S A B C$ with vertex $S$ and base $A B C$; the height of the pyramid $S K$ is 6. Prove that there exists a unique plane intersecting the edges of the base $A B$ and $B C$ at some points $M$ and $N$ such that $|M N|=$ $=7$, touching the sphere at a point equidistant from points $M$ and $N$, and intersecting the extension of the pyramid's height $S K$ beyond point $K$ at some point $D$. Find the length of the segment $S D$.
109. Given the radius of the sphere inscribed in a regular triangular pyramid and the height of the pyramid, it is not difficult to find the side of the base. It is equal to $12, |M K|=|K N|$ (by the condition, the tangents to the sphere from points $M$ and $N$ are equal).

Let $|B M|=x, |B N|=y$. By finding $|M N|$ using the cosine theorem in $\triangle B M N$, and $|M K|$ and $|N K|$ from triangles $B M K$ and $B N K$, we obtain the system of equations

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } - x y = 4 9 } \\
{ x ^ { 2 } - 1 2 x = y ^ { 2 } - 1 2 y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2}-x y=49 \\
(x-y)(x+y-12)=0
\end{array}\right.\right.
$$

This system has the solution $x_{1}=y_{1}=7$. In this case, the distance from $K$ to $M N$ is $4 \sqrt{3}-\frac{7 \sqrt{3}}{2}=\frac{\sqrt{3}}{2}2$, i.e., in this case, the plane passing through $M N$ and tangent to the sphere does not satisfy the condition of the problem.

Answer: $6 \frac{12}{13}$.