## Task Condition

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

$M(1 ; 2 ; 3)$

$\frac{x-0.5}{0}=\frac{y+1.5}{-1}=\frac{z-1.5}{1}$
## 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:

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

Then the equation of the desired plane is:

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

$-y+2+z-3=0$

$-y+z-1=0$

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

We write the parametric equations of the line.

$\frac{x-0.5}{0}=\frac{y+1.5}{-1}=\frac{z-1.5}{1}=t \Rightarrow$

$\left\{\begin{array}{l}x=0.5 \\ y=-1.5-t \\ z=1.5+t\end{array}\right.$

Substitute into the equation of the plane:

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

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

$2 t+2=0$

$t=-1$

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

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

We get:

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

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

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

We get:

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