Example 1.31. Find the coordinates of the point $M$ that bisects the segment of the line

$$
\frac{x-2}{3}=\frac{y+1}{5}=\frac{z-3}{-1}
$$

enclosed between the planes $x o z$ and $x o y$.
Solution. To find the point $A$ of intersection of the line with the $x o z$ plane, we set $y=0$ in the equations of the line. Then we get:

$$
\frac{x-2}{3}=\frac{1}{5}=\frac{z-3}{-1}, \text { or } \quad \begin{aligned}
& \frac{x-2}{3}=\frac{1}{5} \\
& \frac{z-3}{-1}=\frac{1}{5}
\end{aligned}
$$

from which $\quad x=2.6 ; \quad z=2.8$. Thus, $A(2.6 ; 0 ; 2.8)$.

Similarly, setting $z=0$ in the equations of the line, we find the coordinates of the point $B$ of intersection of the line with the $x o y$ plane:

$$
\frac{x-2}{3}=\frac{y+1}{5}=3 \quad \text { or } \quad \begin{aligned}
& \frac{x-2}{3}=3 \\
& \frac{y+1}{5}=3
\end{aligned}
$$

from which $x=11, y=14$, or $B(11 ; 14 ; 0)$.

We determine the coordinates of the point $M$ that divides the segment $A B$ in half:

$$
\begin{gathered}
x_{M}=\frac{x_{A}+x_{B}}{2}=\frac{2.6+11}{2}=6.8 \\
y_{M}=\frac{y_{A}+y_{B}}{2}=\frac{0+14}{2}=7 \\
z_{M}=\frac{z_{A}+z_{B}}{2}=\frac{2.8+0}{2}=1.4
\end{gathered}
$$

Therefore, the coordinates of the desired point $M$ will be: $M(6.8 ; 7 ; 1.4)$.