## Problem Statement

Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.

$M(2; -2; -3)$

$y+z+2=0$
## Solution

Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:

$\vec{s}=\vec{n}=\{0 ; 1 ; 1\}$

Then the equation of the desired line is:

$\frac{x-2}{0}=\frac{y+2}{1}=\frac{z+3}{1}$

Let's find the point $M_{0}$ of intersection of the line and the plane.

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-2}{0}=\frac{y+2}{1}=\frac{z+3}{1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=2 \\
y=-2+t \\
z=-3+t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$(-2+t)+(-3+t)+2=0$

$-2+t-3+t+2=0$

$2 t-3=0$

$t=1.5$

Find the coordinates of the point of intersection of the line and the plane:

$$
\left\{\begin{array}{l}
x=2 \\
y=-2+1.5=-0.5 \\
z=-3+1.5=-1.5
\end{array}\right.
$$

We get:

$$
M_{0}(2 ;-0.5 ;-1.5)
$$

Since $M_{0}$ is the midpoint of the segment $M M^{\prime}$, then

$$
\begin{aligned}
x_{M_{0}} & =\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 2-2=2 \\
y_{M_{0}} & =\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-0.5)-(-2)=1
\end{aligned}
$$

$$
z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot(-1.5)-(-3)=0
$$

We get:

$M^{\prime}(2 ; 1 ; 0)$