## [Coordinate method in space ]

[ equation of a plane $]$

Form the equation of the plane passing through the point $M 0(x 0 ; y 0 ; z 0)$ perpendicular to a non-zero vector $\vec{n}$ $=(a ; b ; c)$.
Point $M(x ; y ; z)$ belongs to the desired plane $\alpha$ if and only if the vector $\overrightarrow{M_{0} M}=(x-x 0 ; y-y 0 ; z-z 0)$ is perpendicular to vector $\vec{n} \cdot$ Therefore,

$$
\begin{aligned}
& M(x ; y ; z) \in \alpha \Leftrightarrow \vec{n} \perp \overrightarrow{M_{0} M} \Leftrightarrow \vec{n} \cdot \overrightarrow{M_{0} M}=0 \Leftrightarrow \\
& \Leftrightarrow a(x-x 0)+b(y-y 0)+c(z-z 0)=0
\end{aligned}
$$

## Answer

$a(x-x 0)+b(y-y 0)+c(z-z 0)=0$