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
$(2,20,24);\;(4,15,23);\;(-2,30,26);\;(0,25,25)$

Long solution
The cartesian equation of the plane is $$\pi: 2x-5y-z-4=0$$
The parametric equations of the line $r$ which is perpendicular to $\pi$ and contains $A=(10,0,20)$ are $$r:\begin{cases}x=10+2t\\y=-5t\\z=20-t\end{cases}$$. This line also contains $B$ so the coordinates of $B$ are $(10+t,-5t,20-t)$ for some $t\in \Bbb R$.
The cartesian equations of $r$ are $$r:\begin{cases} y-5z+100=0 \\x+2z-50=0\end{cases}$$
The line $r$ intersects $\pi$ in the middle point $M$ between $A$ and $B$. Let's find the coordinates of $M$ by solving :$$\begin{cases} 2x-5y-z-4=0\\y-5z+100=0 \\x+2z-50=0\end{cases}$$
Once you've found the coordinates of $M=(x_M,y_M,z_M)$ you solve for $t$ $$||A-M||=||B-M||$$ $$\sqrt{(10-x_M)^2+(-y_M)^2+(20-z_M)^2}=\sqrt{(10+t-x_M)^2+(-5t-y_M)^2+(20-t-z_M)^2}$$
Lastly substitute the value of $t$ you've just found in $(10+t,-5t,20-t)$ to find the coordinates of $B$.
Shorter solution
For each alternative point $P$ that the exercise gives you, calculate the vector $\vec{AP}$ and check if it's parallel to the vector $(2,-5,-1)$ which is perpendicular to the plane. If this is true for more then one of the alternatives calculate the distance of $A$ and $P$ from $\pi$ and check if they're equal. The one alternative that verifies these two conditions ($\vec {AP}$ perpendicular to $\pi$ and $A$ and $P$ at the same distance from $\pi$) is $B$.
