Problem 7. Find the distance $B_{1} H$ from point $B_{1}$ to the line $D_{1} B$, if $B_{1}(5 ; 8 ;-3), D_{1}(-3 ; 10 ;-5)$, $B(3 ; 4 ; 1)$.
Solution. $\quad$ Let $\overline{D_{1} H}=n \overline{D_{1} B}=n(6 ; \quad-6 ; 6)$. Further,

$$
\begin{aligned}
\overline{H B_{1}} & =\overline{D_{1} B_{1}}-\overline{D_{1} H}=(8 ;-2 ; 2)-n(6 ;-6 ; 6)= \\
& =(8-6 n ;-2+6 n ; 2-6 n) .
\end{aligned}
$$

According to the problem, $\overline{D_{1} B} \cdot \overline{H B}_{1}=0$. Therefore,

$$
6(8-6 n)-6(-2+6 n)+6(2-6 n)=\underline{0} .
$$

From this, $n=2: 3$. From equality (1) we have $\overline{H B}_{1}=(4$; $2 ;-2)$

$$
\overline{H B}_{1}^{2}=4^{2}+2^{2}+(-2)^{2}=24 \text { and } \overline{H B}_{1}=2 \sqrt{6}
$$