## Problem Statement

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

$$
\begin{aligned}
& M(1 ; 0 ; 1) \\
& 4 x+6 y+4 z-25=0
\end{aligned}
$$
## 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:

$\vec{s}=\vec{n}=\{4 ; 6 ; 4\}$

Then the equation of the desired line is:

$\frac{x-1}{4}=\frac{y}{6}=\frac{z-1}{4}$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{4}=\frac{y}{6}=\frac{z-1}{4}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+4 t \\
y=6 t \\
z=1+4 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$4(1+4 t)+6(6 t)+4(1+4 t)-25=0$

$4+16 t+36 t+4+16 t-25=0$

$68 t-17=0$

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

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

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

We get:

$$
M_{0}(2 ; 1.5 ; 2)
$$

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

$x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 2-1=3$

$y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot 1.5-0=3$

$z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 2-1=3$

We get:

$M^{\prime}(3 ; 3 ; 3)$