Question

Find an equation of the plane satisfying the following: 1. the plane contains the points (3, -1, 0) and (2, 2, -3). 2. the plane is perpendicular to the plane x-3y+5z=6

Answer 1

if the vector perpendicular to the required plane is $$<a,b,c>$$
according to the given data
$$[\langle 2,2,-3\rangle -\langle 3,-1,0\rangle]\cdot\langle a,b,c\rangle=0$$
$$-a+3b-3c=0 \qquad (1)$$
we know that if the equation of a plane is $$px+qy+rz=c$$
a vector perpendicular to it is $$\langle p,q,r\rangle$$ from that we can say that a vector perpendicular to given equation is $$\langle 1,-3,5\rangle$$ Since this the considered two planes are perpendicular, the vectors perpendicular to each planes are perpendicular to each other. Thus,
$$\langle a,b,c\rangle\cdot\langle1,-3,5 \rangle=0\implies a-3b+5c=0\qquad (2)$$
From (1) and (2) c=0, a=3b

therefore a vector perpendicular to the required plane is $$\langle 3b,b,0\rangle$$ or$$\langle 3,1,0\rangle$$
Now you have a vector perpendicular to the plane and a point on the plane. Now you can find the equation of the plane