Find the point on the z-axis which is equidistant from the points

Consider, C(0,0,z) point which lies on z axis and is equidistant from points A(1, 5, 7) and B(5, 1, -4)

∴ AC = BC

$\sqrt{(0-1)^{2}+(0-5)^{2}+(z-7)^{2}}=\sqrt{(0-5)^{2}+(0-1)^{2}+(z+4)^{2}}$

Squaring both sides,

$(0-1)^{2}+(0-5)^{2}+(z-7)^{2}=(0-5)^{2}+(0-1)^{2}+(z+4)^{2}$

$1+25+z^{2}-14 z+49=25+1+z^{2}+8 z+16$

$-22 z=-33$

Z = 1.5

The point C is (0,0,1.5).