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

**To find:** the 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(1, 2, 3) is PA,

\(\sqrt{(x-1)^2+(y−2)^2+(z-3)^2}\)

The distance between P(x, y, z) and B(3, 2, -1) is PB,

=\(\sqrt{(x-3)^2+(y−2)^2+(z-(-1))^2}\)

= \(\sqrt{(x-3)^2+(y−2)^2+(z+1))^2}\)

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

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

⇒ x^{2}+ 1 – 2x + y^{2 }+ 4 – 4y + z^{2} + 9 – 6z = x^{2}+ 9 – 6x + y^{2} + 4 – 4y + z^{2} + 1 + 2z

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

⇒ 4x – 8z = 0

⇒ 4(x – 2z) = 0

⇒ x – 2z = 0

Hence locus of point P is x – 2z = 0