## Problem Statement

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

$M(2 ; 1 ; 0)$

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

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

Now, let's find the point $M_{0}$ of intersection between the line and the plane.

We write the parametric equations of the line.

$\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}=t \Rightarrow$

$\left\{\begin{array}{l}x=2 \\ y=1+t \\ z=t\end{array}\right.$

Substitute into the equation of the plane:

$(1+t)+t+2=0$

$2 t+3=0$

$t=-1.5$

Now, let's find the coordinates of the intersection point of the line and the plane:

$$
\left\{\begin{array}{l}
x=2 \\
y=1+(-1.5)=-0.5 \\
z=-1.5
\end{array}\right.
$$

We get:

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

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

$x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 2-2=2$

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

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

We get:

$M^{\prime}(2 ;-2 ;-3)$