## Task Condition

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

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

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

Then the equation of the desired plane is:

$$
\begin{aligned}
& 1 \cdot(x-0)-1 \cdot(y-(-3))+1 \cdot(z-(-2))=0 \\
& x-y-3+z+2=0 \\
& x-y+z-1=0
\end{aligned}
$$

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

We write the parametric equations of the line.

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

Substitute into the equation of the plane:

$(1+t)-(-1.5-t)+t-1=0$

$1+t+1.5+t+t-1=0$

$3 t+1.5=0$

$t=-0.5$

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

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

We get:

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

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

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

We get:

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