\[
\left[\begin{array}{l}
\text { Coordinate method in space } \\
{[\text { equation of a plane }]}
\end{array}\right]
\]

Given points \( A(1 ; 0 ; 1), B(-2 ; 2 ; 1), C(2 ; 0 ; 3) \) and \( D(0 ; 4 ;-2) \). Form the equation of the plane passing through point \( B \) and parallel to the plane \( ABC \).
Let's find the coordinates of vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:

$$
\begin{aligned}
& \overrightarrow{A B}=(-2-1 ; 2-0 ; 1-1)=(-3 ; 2 ; 0) \\
& \overrightarrow{A C}=(2-1 ; 0-0 ; 3-1)=(1 ; 0 ; 2)
\end{aligned}
$$

Let $\vec{n}=(a ; b ; c)$ be a non-zero vector perpendicular to the plane $A B C$. Then $\vec{n} \cdot \overrightarrow{A B}=0$ and $\vec{n} \cdot \overrightarrow{A C}=0$, or

$$
\left\{\begin{array}{l}
-3 a+2 b=0 \\
a+2 c=0
\end{array}\right.
$$

Let $c=-1$. Then $a=2, b=\frac{3}{2} a=3$. Since the desired plane is parallel to the plane $A B C$, the vector $\vec{n}$ is also perpendicular to the desired plane. The equation of the plane passing through the point $D(0 ; 4 ;-2)$ and perpendicular to the vector $\vec{n}=(2 ; 3 ;-1)$ is

$$
2 x+3(y-4)-(z+2)=0, \text { or } 2 x+3 y-z-14=0 .
$$

## Answer

$2 x+3 y-z-14=0$.