## Problem Statement

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

$M(3 ; 3 ; 3)$

$\frac{x-1}{-1}=\frac{y-1.5}{0}=\frac{z-3}{1}$
## Solution

Find the equation of the plane that is perpendicular to the given line and passes through point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as the normal vector of the plane:

$\vec{n}=\vec{s}=\{-1 ; 0 ; 1\}$

Then the equation of the desired plane is:

$-1 \cdot(x-3)+0 \cdot(y-3)+1 \cdot(z-3)=0$

$-x+3+z-3=0$

$-x+z=0$

Find the point $M_{0}$ of intersection between the line and the plane.

Write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{-1}=\frac{y-1.5}{0}=\frac{z-3}{1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1-t \\
y=1.5 \\
z=3+t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

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

$-1+t+3+t=0$

$2+2 t=0$

$t=-1$

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

$$
\left\{\begin{array}{l}
x=1-(-1)=2 \\
y=1.5 \\
z=3+(-1)=2
\end{array}\right.
$$

We get:

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

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

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

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

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

We get:

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