## Problem Statement

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

$M(2; -1; 1)$

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

$\vec{s}=\vec{n}=\{1 ;-1 ; 2\}$

Then the equation of the desired line is:

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

Now, let's find the point $M_{0}$ where the line intersects the plane.

We write the parametric equations of the line.

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

Substitute into the equation of the plane:

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

$2+t+1+t+2+4 t-2=0$

$3+6 t=0$

$t=-0.5$

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

$$
\left\{\begin{array}{l}
x=2+(-0.5)=1.5 \\
y=-1-(-0.5)=-0.5 \\
z=1+2(-0.5)=0
\end{array}\right.
$$

We get:
$M_{0}(1.5 ;-0.5 ; 0)$

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

$x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 1.5-2=1$

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

$z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 0-1=-1$

We get:

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