Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Let A (1, 2, 3) & B (3, – 2, – 1)
Let point P be (x, y, z)
Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, – 2, – 1)
i.e. PA = PB
Calculating PA
P ≡ (x, y, z) and A ≡ (1, 2, 3)
Distance PA 
Here,
x1 = x, y1 = y, z1 = z
x2 = 1, y2 = 2, z2 = 3
Distance PA 
Calculating PB
P ≡ (x, y, z) and B ≡ (3, – 2, – 1)
Distance PB 
Here,
x1 = x, y1 = y, z1 = z
x2 = 3, y2 = – 2, z2 = – 1
Distance PB 
Since PA = PB
Squaring both sides, we get –
PA2 = PB2
⇒ (1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (– 2 – y)2 + (– 1 – z)2
⇒ (1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)
= (9 + x2 – 6x) + (4 + y2 + 4y) + (1 + z2 + 2z)
⇒ – 2x – 4y – 6z + 14 = – 6x + 4y + 2z + 14
⇒ 4x – 8y – 8z = 0
⇒ x – 2y – 2z = 0
This is the required equation.