Answer:
the point C is closest to the yz-plane but any of the 3 points lie on this plane
Step-by-step explanation:
the distance of any point (x,y,z) to the yz-plane (0,y,z) is the modulus of the vector d,such that
d= (x,y,z)- (0,y,z) = (x,0,0)
then
|d|= √(x²+0²+0²)= |x|
then the distance is given by the x coordinate of the points
for
A(−2, 0, −6) → |d| = |-2| = 2
B(4, 2, −5) → |d| = |4| = 4
C(1, 3, 2) → |d| = |1| = 1
then the point C is closest to the yz-plane
a point lies on the yz-plane of the distance to it is 0 (|d|=|x|=0) , then any of the 3 points lie on this plane.