## Problem Statement

Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(-2 ; 0 ; 3)$
$2 x-2 y+10 z+1=0$
## 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 use the normal vector of the plane as its direction vector:

$\vec{s}=\vec{n}=\{2 ;-2 ; 10\}$

Then the equation of the desired line is:

$\frac{x+2}{2}=\frac{y}{-2}=\frac{z-3}{10}$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+2}{2}=\frac{y}{-2}=\frac{z-3}{10}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-2+2 t \\
y=-2 t \\
z=3+10 t
\end{array}\right.
\end{aligned}
$$

$2(-2+2 t)-2(-2 t)+10(3+10 t)+1=0$

$-4+4 t+4 t+30+100 t+1=0$

$27+108 t=0$

$t=-\frac{1}{4}$

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

$$
\left\{\begin{array}{l}
x=-2+2 \cdot\left(-\frac{1}{4}\right)=-2.5 \\
y=-2 \cdot\left(-\frac{1}{4}\right)=0.5 \\
z=3+10 \cdot\left(-\frac{1}{4}\right)=0.5
\end{array}\right.
$$

We get:

$$
M_{0}(-2.5 ; 0.5 ; 0.5)
$$

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

$$
\begin{aligned}
& x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot(-2.5)-(-2)=-3 \\
& y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot 0.5-0=1 \\
& z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 0.5-3=-2
\end{aligned}
$$

We get:

$M^{\prime}(-3 ; 1 ;-2)$