## Condition of the problem

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

$$
\begin{aligned}
& M(2 ;-2 ;-3) \\
& \frac{x-1}{-1}=\frac{y+0.5}{0}=\frac{z+1.5}{0}
\end{aligned}
$$
## Solution

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

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

Then the equation of the desired plane is:

$-1 \cdot(x-2)+0 \cdot(y+2)+0 \cdot(z+3)=0$

$-x+2=0$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{-1}=\frac{y+0.5}{0}=\frac{z+1.5}{0}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1-t \\
y=-0.5 \\
z=-1.5
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

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

$-1+t+2=0$

$t=-1$

We find the coordinates of the intersection point between the line and the plane:

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

We get:

$M_{0}(2 ;-0.5 ;-1.5)$

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

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

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

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

We get:

$M^{\prime}(2 ; 1 ; 0)$