88. The edge of a regular tetrahedron is $a$. The plane $\boldsymbol{P}$ passes through the vertex $B$ and the midpoints of the edges $A C$ and $A D$. A sphere touches the lines $A B, A C, A D$ and that part of the plane $P$ which is enclosed within the tetrahedron. Find the radius of the sphere.
88. Let's conduct a section with a plane passing through edge $A B$ and point $L$ - the midpoint of $C D, K$ - the intersection point of plane $P$ and $A L$, the height dropped from $A$ to $B L$ intersects $B K$ at point $N$, and $B L-$ at point $Q$ (Fig. 9). It is not difficult to prove that the center of the sphere lies on the line $A Q$. The center of the sphere can be on the segment $A N$ (point $O$) and on the extension of $A Q$ (point $O_{1}$).

The radius of the first sphere is equal to the radius of the circle tangent to $\boldsymbol{A} \boldsymbol{B}$ and $\boldsymbol{B} K$ and having its center on $A N$. Let's denote it by $x$; $x$ can be found from the relation

$$
\begin{gathered}
S_{B A N}=\frac{1}{2}(|A B|+|B N|) x \\
|B N|=\frac{4}{5}|B K|=\frac{2}{5} \sqrt{2|A B|^{2}+2|B L|^{2}-|A L|^{2}}=\frac{\sqrt{11}}{5} a \\
S_{B A N}=\frac{2}{5} S_{B A L}=\frac{\sqrt{2}}{10} a^{2}
\end{gathered}
$$

Thus, $x=\frac{\sqrt{2} a}{5+\sqrt{11}}$. The radius of the second sphere is found similarly,

$$
\text { Answer: } \frac{\sqrt{2} a}{5 \pm \sqrt{11}}
$$