Answer:
See answer below
Step-by-step explanation:
Since the yz-plane represents all the points (0,y,z) , then the distance of any point (x,y,z) to the yz-plane is given by the difference vector
d= (x,y,z) - (0,y,z) = (x,0,0)
then the distance |d| is
|d|=√ ( x² + 0²+0²) = |x|
then the distance only depends on |x|. Therefore A has the minimum distance to the plane and its distance dA is
dA=|xA| =|-6| = 6
Similarly , the xy- plane represents all the points (x,y,0). Doing the same procedure , the distance is |z| . Then the point that is closest to this plane is B because
dB=|zB| =|-1| = 1
Also , for a point to lie in the xz-plane , has to be of the form (x,y,0) , that is zB=0 ( distance to the plane =0). Since neither A , B or C comply , then any of the points lie in this plane
