question in solid geometry

Question

How to find point A belonging to the plane given by the equation $x+y-2z=3$ as the line segment BA is perpendicular to the plane (where B is given by (2,1,6) ).

Answer 1

You can move orthogonally from $B$ towards the plane, say $\pi:x+y-2z=3$. The direction orthogonal to $\pi$ is given by the vector $v=(1,1,-2)$ if you move from $B$ in the $v$-direction you get points of the form $B_t=(2+t,1+t,6-2t)$ (the first one, for $t=0$ is $B$). You search for the $B_t$ point which is in $\pi$. So sobstitute in the equation of $\pi$ searching for the right $t$. This give you $t=2$ and $A:=B_2=(4,3,2)\in\pi$