## Problem Statement

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

$M(1 ; 2 ; 3)$

$2 x+10 y+10 z-1=0$
## 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 take the normal vector of the plane as its direction vector:

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

Then the equation of the desired line is:

$\frac{x-1}{2}=\frac{y-2}{10}=\frac{z-3}{10}$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{2}=\frac{y-2}{10}=\frac{z-3}{10}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+2 t \\
y=2+10 t \\
z=3+10 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

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

$2+4 t+20+100 t+30+100 t-1=0$

$204 t+51=0$

$t=-0.25$

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

$$
\left\{\begin{array}{l}
x=1+2 \cdot(-0.25)=0.5 \\
y=2+10 \cdot(-0.25)=-0.5 \\
z=3+10 \cdot(-0.25)=0.5
\end{array}\right.
$$

We get:

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

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

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

$y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-0.5)-2=-3$

$z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 0.5-3=-2$

We get:

$M^{\prime}(0 ;-3 ;-2)$