4.14. Find the coordinates of the point $A^{\prime}$, symmetric to the point $A(2,0,2)$ with respect to the plane $4 x+6 y+4 z-50=0$.
Solution. Let's write the equation of the line passing through point $\boldsymbol{A}$ and perpendicular to the given plane:

$$
(x-2) / 4=(y-0) / 6=(z-2) / 4
$$

Convert the equation of the line to parametric form:

$$
\begin{aligned}
& (x-2) / 4=(y-0) / 6=(z-2) / 4=t \\
& x=2+2 t, \quad y=3 t, \quad z=2+2 t
\end{aligned}
$$

Similarly to how it was done in problem 4.12, let's find the coordinates of the intersection point of the found line and the given plane:

$$
\begin{aligned}
& 4 \cdot(2+2 t)+6 \cdot(3 t)+4 \cdot(2+2 t)-50=0, t=1, \\
& x_{P}=4, \quad y_{P}=3, \quad z_{P}=4 .
\end{aligned}
$$

Using the formulas for dividing a segment in a given ratio, we determine the coordinates of point $A^{\prime}$ from the condition that $P$ is the midpoint of segment $A A^{\prime}$:

$$
\begin{aligned}
& x_{P}=\left(x_{A}+x_{A^{\prime}}\right) / 2, y_{P}=\left(y_{A}+y_{A^{\prime}}\right) / 2, z_{P}=\left(z_{A}+z_{A^{\prime}}\right) / 2 \\
& x_{A^{\prime}}=6, y_{A^{\prime}}=6, z_{A^{\prime}}=6
\end{aligned}
$$