Condition of the problem

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

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

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

Then the equation of the desired plane is:

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

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

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

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

We write the parametric equations of the line.

$\frac{x-1.5}{2}=\frac{y}{-1}=\frac{z-2}{1}=t \Rightarrow$

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

Substitute into the equation of the plane:

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

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

$6 t+6=0$

$t=-1$

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

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

We get:

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

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

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

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

We get:

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