## Problem Statement

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

$$
M(-1 ; 0 ;-1)
$$

$2 x+6 y-2 z+11=0$
## 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:

$\vec{s}=\vec{n}=\{2 ; 6 ;-2\}$

Then the equation of the desired line is:

$\frac{x+1}{2}=\frac{y}{6}=\frac{z+1}{-2}$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+1}{2}=\frac{y}{6}=\frac{z+1}{-2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-1+2 t \\
y=6 t \\
z=-1-2 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$2(-1+2 t)+6(6 t)-2(-1-2 t)+11=0$

$-2+4 t+36 t+2+4 t+11=0$

$44 t+11=0$

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

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

$$
\left\{\begin{array}{l}
x=-1+2 \cdot\left(-\frac{1}{4}\right)=-1.5 \\
y=6 \cdot\left(-\frac{1}{4}\right)=-1.5 \\
z=-1-2 \cdot\left(-\frac{1}{4}\right)=-0.5
\end{array}\right.
$$

We get:

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

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

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

$$
\begin{aligned}
& y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-1.5)-0=-3 \\
& z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot(-0.5)-(-1)=0
\end{aligned}
$$

We get:

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