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Q.
The point on the x-axis equidistant from the points $ (4,3,1) $ and $ (-2,\,-6,\,-2) $ is
J & K CETJ & K CET 2009Introduction to Three Dimensional Geometry
Solution:
Let the point on x-axis is $ (x,0,0) $
According to the question,
$ \sqrt{{{(x-4)}^{2}}+{{(0-3)}^{2}}+{{(0-1)}^{2}}} $
$ =\sqrt{{{(x+2)}^{2}}+{{(0+6)}^{2}}+{{(0+2)}^{2}}} $
$ \Rightarrow $ $ {{x}^{2}}+16-8x+9+1={{x}^{2}}+4+4x+36+4 $
$ \Rightarrow $ $ 12x=26-44=-18 $
$ \Rightarrow $ $ x=\frac{-18}{12}=\frac{-3}{2} $
Hence, required point is $ \left( \frac{-3}{2},0,0 \right) $ .