Find the coordinates of the point which is equidistant from the four points *O*(0, 0, 0), *A*(2, 0, 0), *B*(0, 3, 0) and *C*(0, 0, 8).

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#### Solution

Let *P* (*x*, *y*, *z*) be the required point which is equidistant from the points *O*(0,0,0), *A*(2,0,0)*B*(0,3,0) and *C*(0,0,8)

Then,*OP* = *AP *

*\[\Rightarrow O P^2 = A P^2\]*

\[\therefore x^2 + y^2 + z^2 = \left( x - 2 \right)^2 + y^2 + z^2 \]

\[ \Rightarrow x^2 = \left( x - 2 \right)^2 \]

\[ \Rightarrow x^2 = x^2 - 4x + 4\]

\[ \Rightarrow 4x = 4\]

\[ \Rightarrow x = \frac{4}{4}\]

\[ \therefore x = 1\]

Similarly, we have:

\[ \Rightarrow x^2 = \left( x - 2 \right)^2 \]

\[ \Rightarrow x^2 = x^2 - 4x + 4\]

\[ \Rightarrow 4x = 4\]

\[ \Rightarrow x = \frac{4}{4}\]

\[ \therefore x = 1\]

Similarly, we have:

*OP*=*BP**\[\Rightarrow O P^2 = B P^2 \]*

\[\therefore x^2 + y^2 + z^2 = x^2 + \left( y - 3 \right)^2 + z^2 \]

\[ \Rightarrow y^2 = y^2 - 6y + 9\]

\[ \Rightarrow 6y = 9\]

\[ \Rightarrow y = \frac{9}{6}\]

\[ \therefore y = \frac{3}{2}\]

\[\therefore x^2 + y^2 + z^2 = x^2 + \left( y - 3 \right)^2 + z^2 \]

\[ \Rightarrow y^2 = y^2 - 6y + 9\]

\[ \Rightarrow 6y = 9\]

\[ \Rightarrow y = \frac{9}{6}\]

\[ \therefore y = \frac{3}{2}\]

Similarly, we also have:

*OP *= *CP*

*\[\Rightarrow O P^2 = C P^2\]*

*\[\Rightarrow x^2 + y^2 + z^2 = x^2 + y^2 + \left( z - 8 \right)^2 \]\[ \Rightarrow z^2 = z^2 - 16z + 64\]\[ \Rightarrow 16z = 64\]\[ \Rightarrow z = \frac{64}{16}\]\[ \therefore z = 4\]*

Thus, the required point is P \[\left( 1, \frac{3}{2}, 4 \right)\]

Is there an error in this question or solution?

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