$\begin{aligned} & {\left[\begin{array}{l}\text { Method of coordinates in space } \\ \text { [Distance between two points. Equation of a sphere }]\end{array}\right]}\end{aligned}$

Find the distance from the point $M 0(x 0 ; y 0 ; z 0)$ to the plane $A x+B y+C z+D=0$.
Through the given point $M 0$ we draw a line perpendicular to the given plane. As the direction vector of this line, we take the vector $\vec{m}=(A ; B ; C)$. Then the parametric equations of the line are

$$
\left\{\begin{array}{l}
x-x_{0}=A t \\
y-y_{0}=B t \\
z-z_{0}=C t
\end{array}\right.
$$

Substituting $x, y$ and $z$, expressed from this system in terms of $t$, into the equation of the plane, we find the value of $t$ for which the point $M$ belongs to the given plane:

$$
A(x 0+A t)+B(y 0+B t)+C(z 0+C t)+D=0 \Leftrightarrow(A 2+B 2+C 2) t=-(A x 0+B y 0+C z 0+D) \Leftrightarrow
$$

$$
\Leftrightarrow t=-\frac{A x_{0}+B_{k_{0}}+C_{20}+D}{A^{2}+B^{2}+C^{2}} .
$$

The distance $\rho$ from the point $M 0$ to the given plane is equal to the distance between the points $M 0$ and the found point of intersection of the drawn line with the given plane, i.e.,

$$
\begin{aligned}
\rho & =M 0 M=\sqrt{\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}+\left(z-z_{0}\right)^{2}}= \\
& =\sqrt{A^{2} t^{2}+B^{2} t^{2}+C^{2} t^{2}}=|t| \sqrt{A^{2}+B^{2}+C^{2}}= \\
& =\frac{\left|A x_{0}+B_{X}+C_{z_{0}}+D\right|}{A^{2}+B^{2}+C^{2}} \cdot \sqrt{A^{2}+B^{2}+C^{2}}=\frac{\left|A x_{0}+B_{X}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}} .
\end{aligned}
$$

## Answer

$\rho=\frac{\left|A x_{0}+B_{20}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$.

Note: There are a few typographical errors in the original text, such as $B_{X}$ and $B_{20}$, which should be $B y_{0}$. The corrected formula should be:

$$
\rho=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}
$$