Since x-coordinate of every point in yz-plane is zero. Let P (0, y, z) be a point on the yz-plane such that PA = PB = PC.
Now PA = PB
(0 – 2)² + (y – 0)² + (z – 3⁾² = (0 – 0)² + (y – 3)² + (z – 2)²

i.e. z – 3y = 0

and PB = PC

⇒ y² + 9 – 6y + z² + 4 – 4z = y² + z² + 1 – 2z ,

i.e. 3y + z = 6 Simplifying the two equating, we get y = 1, z = 3
Here, the coordinate of the point P are (0, 1, 3).

1Thank You

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