The midpoint of the two given points is (-1+2)/2,(1+7)/2,(0+3)/2)=P(0.5,4,1.5), which lies on the plane. The two given points have to be perpendicular to the plane. Their position vectors in relation to the origin O(0,0,0) are <-1,1,0> and <2,7,3> and the normal vector n=<2-(-1),7-1,3-0>=<3,6,3>.
OP=<0.5,4,1.5>, and if we take a general point on the plane Q(x,y,z), so that OQ=<x,y,z>:
OP+PQ=OQ, and PQ=OQ-OP=<x-0.5,y-4,z-1.5>. Also:
n.PQ=0=3(x-0.5)+6(y-4)+3(z-1.5). 3x+6y+3z-(1.5+24+4.5)=0=3x+6y+3z-30 is the equation of the plane. This reduces to x+2y+z=10.
However, the plane can be equidistant from the given points of the plane lies, not between them, but all around them (hence two planes) an equal distance away. Imagine these points to be the centres of two spheres of the same radius, or a cylinder with axis passing through the points, then the plane will be tangential to both spheres and the cylinder. And we are told that the plane passes through the y-axis (assumedly along its whole length). The plane derived above doesn't do this.
The plane has to be parallel to the line joining the given points, so it shares its normal with this line. This line is given by the difference between the position vectors of the given points: v=<3,6,3> (this was n earlier). Any two points on the plane must lie on a line parallel to v. (Previously the plane was perpendicular to v.) If the plane is to pass through the y-axis then the plane lies on top of the two hypothetical spheres or cylinder or the two spheres or cylinder sit on the plane. If the plane passes through the y-axis then it's perpendicular to the other two axes.
More to follow...