Find an equation of the plane.
The plane through the origin and the points $ (3, -2, 1) $ and $ (1, 1, 1) $
$3 x+2 y-5 z=0$
Hello. So the question is taken from vectors and geometry of the space. And in discussion we have to find the equation of plane that passes to three point origin three minus 21 and 111. So we need to find an equation of the plane that passes 20003 minus 21 and 1 1. So let me write equation of clean passes to Origin 3 -2 1 And 111 can be. Have you done as mm Yeah yeah X minus zero by minus zero. Said minus zero. So in second time three minus zero minus two minus 01 minus zero and third one is one minus zero one. Miner's. You let me right here in order to understand it 3 0 -2 0 1 0 & 1 0. 1 0, 1 0. So if we saw that is equal to zero. So if we solve the situation we get x into minus two minus well okay minus y into three minus one plus that. Into three plus two Which is equal to zero. So equation will be -3 x minus two. Y plus five is equal to zero. So taking minus through to the right hand side we get three x plus two y minus five that is equal to zero which is the required equation of plane that is passing through the required three point. Hope this clears without and thank