## Task Condition

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

$M(1 ; 1 ; 1)$

$$
\frac{x-2}{1}=\frac{y+1.5}{-2}=\frac{z-1}{1}
$$
## 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:

$\vec{n}=\vec{s}=\{1 ;-2 ; 1\}$

Then the equation of the desired plane is:

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

$x-1-2 y+2+z-1=0$

$x-2 y+z=0$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-2}{1}=\frac{y+1.5}{-2}=\frac{z-1}{1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=2+t \\
y=-1.5-2 t \\
z=1+t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$(2+t)-2(-1.5-2 t)+(1+t)=0$

$2+t+3+4 t+1+t=0$

$6+6 t=0$

$t=-1$

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

$$
\left\{\begin{array}{l}
x=2+(-1)=1 \\
y=-1.5-2 \cdot(-1)=0.5 \\
z=1+(-1)=0
\end{array}\right.
$$

We get:

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

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

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

We get:

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