## Problem Statement

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

$M(1 ; 1 ; 1)$
$x+4 y+3 z+5=0$
## 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:

$\vec{s}=\vec{n}=\{1 ; 4 ; 3\}$

Then the equation of the desired line is:

$\frac{x-1}{1}=\frac{y-1}{4}=\frac{z-1}{3}$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y-1}{4}=\frac{z-1}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+t \\
y=1+4 t \\
z=1+3 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$$
\begin{aligned}
& (1+t)+4(1+4 t)+3(1+3 t)+5=0 \\
& 1+t+4+16 t+3+9 t+5=0 \\
& 26 t+13=0 \\
& t=-0.5
\end{aligned}
$$

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

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

We get:

$M_{0}(0.5 ;-1 ;-0.5)$

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

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

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

We get:

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