## Problem Statement

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

$M(3 ; 3 ; 3)$

$8 x+6 y+8 z-22=0$
## 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:

$\vec{s}=\vec{n}=\{8 ; 6 ; 8\}$

Then the equation of the desired line is:

$\frac{x-3}{8}=\frac{y-3}{6}=\frac{z-3}{8}$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-3}{8}=\frac{y-3}{6}=\frac{z-3}{8}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=3+8 t \\
y=3+6 t \\
z=3+8 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$8(3+8 t)+6(3+8 t)+8(3+8 t)-22=0$

$24+64 t+18+48 t+24+64 t-22=0$

$176 t+44=0$

$t=-0.25$

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

$$
\left\{\begin{array}{l}
x=3+8 \cdot(-0.25)=1 \\
y=3+6 \cdot(-0.25)=1.5 \\
z=3+8 \cdot(-0.25)=1
\end{array}\right.
$$

We get:

$$
M_{0}(1 ; 1.5 ; 1)
$$

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

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

We get:

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