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.