Condition of the problem

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

$$
\begin{aligned}
& M(3 ;-3 ;-1) \\
& \frac{x-6}{5}=\frac{y-3.5}{4}=\frac{z+0.5}{0}
\end{aligned}
$$
## Solution

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

$\vec{n}=\vec{s}=\{5 ; 4 ; 0\}$

Then the equation of the desired plane is:

$5 \cdot(x-3)+4 \cdot(y-(-3))+0 \cdot(z-(-1))=0$

$5 x-15+4 y+12=0$

$5 x+4 y-3=0$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-6}{5}=\frac{y-3.5}{4}=\frac{z+0.5}{0}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=6+5 t \\
y=3.5+4 t \\
z=-0.5
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$5(6+5 t)+4(3.5+4 t)-3=0$

$30+25 t+14+16 t+3=0$

$41+41 t=0$

$t=-1$

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

$$
\left\{\begin{array}{l}
x=6+5 \cdot(-1)=1 \\
y=3.5+4 \cdot(-1)=-0.5 \\
z=-0.5
\end{array}\right.
$$

We get:

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

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

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

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

We get:

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