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)

PA = PB

Firstly let us calculate

Calculating PA

P ≡ (x, y, z) and A ≡ (1, 2, 3)

By using the formula,

Distance PA = 

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x₁ = x, y₁ = y, z₁ = z

x₂ = 1, y₂ = 2, z₂ = 3

Distance PA 

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

Calculating PB

P ≡ (x, y, z) and B ≡ (3, 2, – 1)

By using the formula,

Distance PB = 

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x₁ = x, y₁ = y, z₁ = z

x₂ = 3, y₂ = 2, z₂ = – 1

Distance PB = 

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

Since PA = PB

Square on both the 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 – 8z = 0

x – 2z = 0

The required equation is x – 2z = 0