123. In a regular triangular pyramid $S A B C$ ( $S$ - vertex), point $E$ is the midpoint of the apothem of face $S B C$, and points $F, L$, and $M$ lie on edges $A B, A C$, and $S C$ respectively, such that $|A L|=\frac{1}{10}|A C|$. It is known that $\boldsymbol{E F L M}$ is an isosceles trapezoid and the length of its base $\boldsymbol{E} \boldsymbol{F}$ is $\sqrt{7}$. Find the volume of the pyramid.
123. Let the side of the base of the pyramid be $a$, and the lateral edge be $b$. Draw a plane through $F E$ parallel to $A S C$, and denote by $K$ and $N$ the points of intersection of this plane with $B C$ and $S B$. Since $E$ is the midpoint of the apothem of the face $S C B$, we have $|A F|=|C K| = a / 4, |S N|=b / 4, |K E|=2|E N|$.

Draw a line through $L$ parallel to $A S$, and denote by $P$ its point of intersection with $S C$. We will have $|S P|=0.1 b$. Triangles $L P C$ and $F N K$ are similar, their corresponding sides are parallel, and moreover, $L M$ and $F E$ are also parallel, i.e., $|P M| / |M C| = |N E| / |E K| = 1 / 2$, hence $|S M| = 0.4 b$.

Now we find:

$$
|L F|^{2}=\frac{19}{400} a^{2}, \quad |M E|^{2}=\frac{15}{400} a^{2}+\frac{1}{100} b^{2}
$$

From the condition $|L F|=|M E|$, we get $a=b$. Triangle $F N K$ is equilateral with side $\frac{3}{4} a, |F E|^{2}=\frac{7}{16} a^{2}=7$. Therefore, $a=b=4$.

Answer: $\frac{16}{3} \sqrt{2}$.