62*. Given two points: $A_{1}\left(x_{1}, y_{1}, z_{1}\right)$ and $A_{2}\left(x_{2}, y_{2}, z_{2}\right)$. Point $P$ defines two vectors: $\overrightarrow{P A}_{1}=\vec{q}_{1}$ and $\overrightarrow{P A}_{2}=\vec{q}_{2}$. What does the set of points $P$ represent, for which the condition $\vec{q}_{1} \vec{q}_{2}=0$ is satisfied?
62. According to the problem, we have: $\vec{q}_{1}=\vec{a}_{1}-\vec{r}, \vec{q}_{2}=\vec{a}_{2}-\vec{r}$. Here $\vec{r}=\overrightarrow{O P}, \vec{a}_{1}=\overrightarrow{O A}_{1}, \vec{a}_{2}=\overrightarrow{O A}_{2}$

$$
\vec{a}_{1} \vec{q}_{2}=\left(\overrightarrow{a_{1}}-\vec{r}\right)\left(\overrightarrow{a_{2}}-\vec{r}\right)=\vec{a}_{1} \vec{a}_{2}-\left(\vec{a}_{1}+\vec{a}_{2}\right) \vec{r}+\vec{r}^{2}=0
$$

Transitioning to coordinates, we get:

$$
\begin{gathered}
r^{2}=x^{2}+y^{2}+z^{2},\left(\overrightarrow{a_{1}}+\overrightarrow{\left.a_{2}\right) r}=\left(x_{1}+x_{2}\right) x+\left(y_{1}+y_{2}\right) y+\right. \\
\quad+\left(z_{1}+z_{2}\right) z \\
\vec{a}_{1} \vec{a}_{2}=x_{1} x_{2}+y_{1} y_{2}+z_{1} z_{2}
\end{gathered}
$$

For brevity, let: $x_{1}+x_{2}=2 x_{0}, \quad y_{1}+y_{2}=2 y_{0}$, $z_{1}+z_{2}=2 z_{0}$. Then our equation will take the following form:

$$
x^{2}+y^{2}+z^{2}-2 x_{0} x-2 y_{0} y-2 z_{0} z+x_{1} x_{2}+y_{1} y_{2}+z_{1} z_{2}=0
$$

To obtain the canonical equation of the sphere from this, add and subtract the sum $x_{0}^{2}+y_{0}^{2}+z_{0}^{2}$:

$$
\begin{aligned}
& \left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}+\left(z-z_{0}\right)^{2}+x_{1} x_{2}+y_{1} y_{2}+z_{1} z_{2}- \\
& \quad-\left(\frac{x_{1}+x_{2}}{2}\right)^{2}-\left(\frac{y_{1}+y_{2}}{2}\right)^{2}-\left(\frac{z_{1}+z_{2}}{2}\right)^{2}=0
\end{aligned}
$$

Transform the last terms of the equation as follows:

$$
\begin{gathered}
\frac{4 x_{1} x_{2}+4 y_{1} y_{2}+4 z_{1} z_{2}-x_{1}^{2}-x_{2}^{2}-2 x_{1} x_{2}-y_{1}^{2}-y_{2}^{2}-2 y_{1} y_{2}-z_{1}^{2}-z_{2}^{2}-2 z_{1} z_{2}}{4}= \\
=-\left(\frac{x_{1}-x_{2}}{2}\right)^{2}-\left(\frac{y_{1}-y_{2}}{2}\right)^{2}-\left(\frac{z_{1}-z_{2}}{2}\right)^{2}=-R^{2}
\end{gathered}
$$

Here $R$ denotes half the distance $\left|A_{1} A_{2}\right|$. Finally, we obtain the equation of the sphere: $\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}+\left(z-z_{0}\right)^{2}-R^{2}=0$. This sphere has the segment $A_{1} A_{2}$ as its diameter.