What are coordinates of the point equidistant from the points $$\left( {a,\,0,\,0} \right),\,\left( {0,\,a,\,0} \right),\,\left( {0,\,0,\,a} \right)$$       and $$\left( {0,\,0,\,0} \right)\,?$$

Solution :
Let the point $$A\left( {x,\,y,\,z} \right)$$   is equidistant from the points $$B\left( {a,\,0,\,0} \right),\,C\left( {0,\,a,\,0} \right),\,D\left( {0,\,0,\,a} \right)$$       and $$E\left( {0,\,0,\,0} \right).$$
$$\eqalign{ & {\text{Hence,}} \cr & \Rightarrow {\left( {x - a} \right)^2} + {y^2} + {z^2} = {x^2} + {\left( {y - a} \right)^2} + {z^2} \cr & \Rightarrow {\left( {x - a} \right)^2} + {y^2} + {z^2} = {x^2} + {y^2} + {\left( {z - a} \right)^2} \cr & \Rightarrow {\left( {x - a} \right)^2} + {y^2} + {z^2} = {x^2} + {y^2} + {z^2} \cr & \Rightarrow {\left( {x - a} \right)^2} + {y^2} + {z^2} = {x^2} + {\left( {y - a} \right)^2} + {z^2} \cr & \Rightarrow {x^2} + {a^2} - 2ax + {y^2} + {z^2} = {x^2} + {y^2} + {a^2} - 2ay + {z^2} \cr & \Rightarrow - 2ax = - 2ay \cr & \Rightarrow ax = ay \cr & \Rightarrow x = y \cr & {\text{Similarly,}} \cr & ay = az \cr & \Rightarrow y = z \cr & \Rightarrow x = y = z \cr & \therefore \,{\left( {x - a} \right)^2} + {x^2} + {x^2} = {x^2} + {x^2} + {x^2} \cr & \Rightarrow {x^2} + {a^2} - 2ax + {x^2} + {x^2} = 3{x^2} \cr & \Rightarrow {a^2} = 2ax \cr & \Rightarrow x = \frac{a}{2} \cr & \therefore \,{\text{Point is }}\left( {\frac{a}{2},\,\frac{a}{2},\,\frac{a}{2}} \right) \cr} $$