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,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = 1, y_{2} = 2, z_{2} = 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,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = 3, y_{2} = 2, z_{2} = – 1

Distance PB =

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

Since PA = PB

Square on both the sides, we get

PA^{2} = PB^{2}

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

(1 + x^{2} – 2x) + (4 + y^{2} – 4y) + (9 + z^{2} – 6z)

(9 + x^{2} – 6x) + (4 + y^{2} – 4y) + (1 + z^{2} + 2z)

– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14

4x – 8z = 0

x – 2z = 0

The required equation is x – 2z = 0

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