Problem 10. Find the coordinates of the point $M(x ; y ; z)$, symmetric to the point $B_{1}(5 ; 8 ;-3)$ with respect to the line defined by the points $B(3 ; 4 ; 1)$ and $A_{1}(1 ; 6 ;-7)$ (Fig. 1).
First solution. $\overline{B H}=0.5 \overline{B B}_{1}+0.5 \overline{B M} ; \overline{B H}=$ $=k \overline{B A_{1}}$.

\[
\begin{gathered}
k \overline{B A_{1}}=0.5 \overline{B B_{1}}+0.5 \overline{B M} \\
\overline{B_{1} M} \cdot \overrightarrow{B A_{1}}=0
\end{gathered}
\]

From equations (1) and (2) we get:

$k(-2 ; 2 ;-8)=0.5(2 ; 4 ;-4)+0.5(x-3 ; y-4 ; z-1)$; $(x-5 ; y-8 ; z+3) \cdot(-2 ; 2 ;-8)=0$.

This system of equations is equivalent to the following system:

\[
\left\{\begin{array}{l}
-2 k=1+0.5(x-3) \\
-8 k=-2+0.5(z-1) \\
2 k=2+0.5(y-4) \\
-2(x-5)+2(y-8)-8(z+3)=0
\end{array}\right.
\]

Solving this system of equations, we get $M(-1 ; 2 ;-3)$.

Second solution. Let $K$ be a point on the line $B A_{1}$ such that the quadrilateral $B B_{1} M K$ is a rhombus. If $\overline{B K}=k \overline{B A_{1}}=(-2 k ; 2 k ;-8 k)$, then $K(3-2 k : 4+2 k ; 1-8 k)$ (Fig. 1). Therefore,

\[
\overline{B_{1} K}=(-2-2 k ;-4+2 k ; 4-8 k) ; \overline{B_{1} B}=(-2 ;-4 ; 4)
\]

Since $\overline{B_{1} K^{2}}=\overline{B_{1} B^{2}}$, we have

\[
\begin{gathered}
(-2-2 k)^{2}+(-4+2 k)^{2}+(4-8 k)^{2}= \\
(-2)^{2}+(-4)^{2}+4^{2}
\end{gathered}
\]

From this, $k=0$ or $k=1$. Now it is clear that $K(1 ; 6 ;-7)$. The point $H$ is the midpoint of the segment $B K$. Therefore, $H(2 ; 5 ;-3)$ and $M(-1 ; 2 ;-3)$.