Answer
$14x -6y-10z = 9$, which is the equation of a plane that passes through the midpoint of points A and B and is perpendicular to the line passing through them.
Work Step by Step
We represent the set of points with a general point $P$ with coordinates $(x, y, z)$.
Using the formula for distance, the distance from $P$ to $A(-1, 5, 3)$ is:
$|PA| =\sqrt {(x+1)^{2} + (y-5)^{2} + (z-3)^{2}}$
And the distance from $P$ to $B(6, 2, -2)$ is:
$|PB| =\sqrt {(x-6)^{2} + (y-2)^{2} + (z+2)^{2}}$.
Now, for the desired set of points, these distances are equal. Thus we can write:
$\sqrt {(x+1)^{2} + (y-5)^{2} + (z-3)^{2}} = \sqrt {(x-6)^{2} + (y-2)^{2} + (z+2)^{2}}$
Squaring both sides, expanding, and combining like terms, we get:
$\sqrt {(x+1)^{2} + (y-5)^{2} + (z-3)^{2}} = \sqrt {(x-6)^{2} + (y-2)^{2} + (z+2)^{2}}$
$(x+1)^{2} + (y-5)^{2} + (z-3)^{2} = (x-6)^{2} + (y-2)^{2} + (z+2)^{2}$
$x^{2}+2x+1 + y^{2}-10y+25 + z^{2}-6z+9 = x^{2}-12x+36 + y^{2}-4y+4 + z^{2}+4z+4$
$14x -6y-10z = 9$
This is the equation of a plane that passes right in the middle of points A and B, and is perpendicular to the line that passes through them.
