The point in XY-plane which is equidistant from three points A (2, 0, 3), B (0, 3, 2) and C (0, 0, 1) has the coordinates

A
$(2, 0, 8)$
B
$(0, 3, 1)$
C
$(3, 2, 0)$
D
$(3, 2, 1)$

Solution:

We know that z-coordinate of every point on xy-plane us zero. So, let P(x,y,0) be a point on xy-plane such that PA=PB=PC.
Now, PA=PB
$\begin{array}{l} \Rightarrow P A^{2} = P B^{2} \\ \Rightarrow (x - 2)^{2} + (y - 0)^{2} + (0 - 3)^{2} \end{array}$
$= (x - 0)^{2} + (y - 3)^{2} + (0 - 2)^{2}$
$\Rightarrow 4 x - 6 y = 0 \Rightarrow 2 x - 3 y = 0$ …(i)
and PB=PC
$\begin{array}{l} \Rightarrow P B^{2} = P C^{2} \\ \Rightarrow (x - 0)^{2} + (y - 3)^{2} + (0 - 2)^{2} \\ \equiv (x - 0)^{2} + (y - 0)^{2} + (0 = 1)^{2} \end{array}$

$\Rightarrow - 6 y + 12 = 0 \Rightarrow y = 2$
Putting $y = 2$ in (i), we obtain $x = 3$
Hence, the required point is $(3, 2, 0)$ .

The point in XY-plane which is equidistant from three points $A (2, 0, 3)$ , $B (0, 3, 2)$ and $C (0, 0, 1)$ has the coordinates

Solution:

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