114. In the pyramid $S A B C$, a line intersecting the edges $A C$ and $B S$ and perpendicular to them passes through the midpoint of the edge $B S$. The face $A S B$ is equal in area to the face $B S C$, and the area of the face $A S C$ is twice the area of the face $B S C$. Inside the pyramid, there is a point $M$, the sum of the distances from which to the vertices $B$ and $S$ is equal to the sum of the distances to all the faces of the pyramid. Find the distance from point $M$ to vertex $B$, if $|A C|=\sqrt{6},|B S|=1$.
114. From the fact that a line perpendicular to the edges $AC$ and $BS$ passes through the midpoint of $BS$, it follows that the faces $ACB$ and $ACS$ are equal in area.

Let $S_{ASB} = S_{BSC} = Q$, then $S_{ACB} = S_{ACS} = 2Q$. Denote by $A_1, B_1, C_1, S_1$ the projections of $M$ onto the faces $BCS, ACS, ABS$, and $ACC$ respectively, $h_A, h_B, h_C, h_S$ - the heights dropped onto these faces, $V$ - the volume of the pyramid. Then we have

$$
\left|M A_1\right| + 2\left|M B_1\right| + \left|M C_1\right| + 2\left|M S_1\right| = \frac{3 V}{Q}
$$

But, by the condition, $\quad|M B| + |M S| = \left|M A_1\right| + \left|M B_1\right| + \left|M C_1\right| + \left|M S_1\right|$. From the last two equalities, it follows:

$$
|M B| + \left|M B_1\right| + |M S| + \left|M S_1\right| = \frac{3 V}{Q}
$$

$H_0$

$$
V = \frac{1}{3} h_S \cdot 2Q = \frac{1}{3} h_B \cdot 2Q = \frac{Q}{3} (h_B + h_S)
$$

Therefore, $\quad|M B| + \left|M B_1\right| + |M S| + \left|M S_1\right| = h_B + h_S$. On the other hand, $|M B| + \left|M B_1\right| \geqslant h_B, |M S| + \left|M S_1\right| \geqslant h_S$. Thus, $|M B| + \left|M B_1\right| = h_B, |M S| + \left|M S_1\right| = h_S$, and the heights dropped from $B$ and $S$ intersect at point $M$, and the edges $AC$ and $BS$ are perpendicular.

From the conditions of the problem, it also follows that the common perpendicular to $AC$ and $BS$ also bisects $AC$. Let $F$ be the midpoint of $AC$, $E$ be the midpoint of $BS$. Denote $|FE| = x$. Then

$$
\begin{gathered}
Q = S_{ASB} = \frac{1}{2} |SB| \cdot |AE| = \frac{1}{2} \sqrt{x^2 + \frac{3}{2}}, \\
2Q = S_{ACB} = \frac{\sqrt{6}}{2} \sqrt{x^2 + \frac{1}{4}}.
\end{gathered}
$$

We obtain the equation $\frac{\sqrt{6}}{2} \sqrt{x^2 + \frac{1}{4}} = \sqrt{x^2 + \frac{3}{2}}$, from which $x = \frac{3}{2}$. Considering the isosceles triangle $BFS$, where $|BS| = 1, |BF| = |FS|$, the height $|FE| = \frac{3}{2}, M$ is the point of intersection of the heights, we find $|BM| = |SM| = \frac{\sqrt{10}}{6}$.