**Given:** Points are A(3, 4, -5) and B(-2, 1, 4)

**To find: **the equation of the set of the points, i.e. locus of points which are equidistant from the given points

Let the required point P(x, y, z)

According to the question:

PA = PB

⇒ PA^{2} = PB^{2}

**Formula used: **

The distance between any two points (a, b, c) and (m, n, o) is given by,

\(\sqrt{(a−m)^2+(b−n)^2+(c−o)^2}\)

Therefore,

The distance between P(x, y, z) and A(3, 4, -5) is PA,

=\(\sqrt{(x-3)^2+(y−4)^2+(z-(-5))^2}\)

=\(\sqrt{(x-3)^2+(y-4)^2+(z+5)^2}\)

The distance between P(x, y, z) and B(-2, 1, 4) is PB,

=\(\sqrt{(x-(-2))^2+(y−1)^2+(z-4)^2}\)

= \(\sqrt{(x+2)^2+(y−1)^2+(z-4)^2}\)

As PA^{2} = PB^{2}

(x – 3)^{2}+ (y – 4)^{2} + (z + 5)^{2} = (x + 2)^{2} + (y – 1)^{2} + (z – 4)^{2}

⇒ x^{2}+ 9 – 6x + y^{2} + 16 – 8y + z^{2} + 25 + 10z = x^{2}+ 4 + 4x + y^{2} + 1 – 2y + z^{2} + 16 – 8z

⇒ x^{2}+ 9 – 6x + y^{2} + 16 – 8y + z^{2} + 25 + 10z – x^{2}– 4 – 4x – y^{2} – 1 + 2y – z^{2} – 16 + 8z = 0

⇒ – 6x – 6y + 18z + 29 = 0

⇒ 6x + 6y – 18z – 29 = 0

Hence locus of point P is 6x + 6y – 18z – 29 = 0