<s>[INST] Symmetrical point with respect to a plane I have the point $A = (10,0,20)$, and I want the coordinates of its symmetrical point $B$ with respect to the plane $\pi = \begin{cases} x=2+3\alpha+\beta\\y=\alpha\\z=\alpha + 2\beta \end{cases}$, how can I do that? Look at the image:
image
These are the question's alternatives
$(2,20,24);\;(4,15,23);\;(-2,30,26);\;(0,25,25)$
 [/INST] An equation of the plane is
\begin{equation}
x = 2 + 3 y + (z-y)/2 \Longleftrightarrow 2 x - 5 y -z -4 = 0
\end{equation}
Let $A = (x_0,y_0,z_0) = (10, 0, 20)$ and $B = (x_1, y_1, z_1)= A + t (2, -5, -1)$ because the vector $(2, -5, -1)$ is perpendicular to the plane. We have $2 x_0-5 y_0 -z_0 - 4 = -4$, hence we need $2 x_1-5 y_1 -z_1 - 4 = +4$ in order to have the midpoint $(A + B)/2$ on the plane. This implies $t = 8/30 = 4/ 15$, hence
\begin{equation}
x_1 = 10 + \frac{8}{15}=\frac{158}{15}\qquad
y_1 = -\frac{20}{15}\qquad z_1 = 20 -\frac{4}{15} = \frac{296}{15}
\end{equation}
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