## Task Condition

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

$M(0; -3; -2)$

$2x + 10y + 10z - 1 = 0$
## Solution

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

$\vec{s}=\vec{n}=\{2 ; 10 ; 10\}$

Then the equation of the desired line is:

$\frac{x}{2}=\frac{y+3}{10}=\frac{z+2}{10}$

Let's find the point $M_{0}$ of intersection of the line and the plane.

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x}{2}=\frac{y+3}{10}=\frac{z+2}{10}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=2 t \\
y=-3+10 t \\
z=-2+10 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$2(2 t)+10(-3+10 t)+10(-2+10 t)-1=0$

$4 t-30+100 t-20+100 t-1=0$

$204 t-51=0$

$t=0.25$

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

$$
\left\{\begin{array}{l}
x=2 \cdot 0.25=0.5 \\
y=-3+10 \cdot 0.25=-0.5 \\
z=-2+10 \cdot 0.25=0.5
\end{array}\right.
$$

We get:

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

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

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

We get:

$M^{\prime}(1 ; 2 ; 3)$