Equation in 3d-Space Consider we have 3 fixed points A,B and C also,We have this Equation :
$MA^2+MB^2+MC^2=30$. 
What does this Equation represent in 3d-Space,( where  M is such a Non-fixed point in that space)?
Suppose $A(a1,b1,b1),B(a2,b2,c2),C(a3,b3,c3),M(x,y,z)$
any help will be appreciated.
 A: Given $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3),M(x,y,z)$ and $r>0$.
What subset of $\mathbb{R}^3$ is specified by the equation
\begin{equation} |MA|^2+|MB|^2+|MC|^2=r^2 \tag{1}\end{equation}
\begin{eqnarray}
|MA|^2&=&(x-x_1)^2+(y-y_1)^2+(z-z_1)^2\\
|MB|^2&=&(x-x_2)^2+(y-y_2)^2+(z-z_2)^2\\
|MC|^2&=&(x-x_3)^2+(y-y_3)^2+(z-z_3)^2
\end{eqnarray}
So we can re-write equation $(1)$ as
\begin{eqnarray}
&&3x^2-2x(x_1+x_2+x_3)+(x_1^2+x_2^2+x_3^2)\\
&+&3y^2-2y(y_1+y_2+y_3)+(y_1^2+y_2^2+y_3^2)\\
&+&3z^2-2z(z_1+z_2+z_3)+(z_1^2+z_2^2+z_3^2)\\
&=&r^2
\end{eqnarray}
Define
\begin{eqnarray}
\bar{x}&=&\frac{x_1+x_2+x_3}{3}\\
\bar{y}&=&\frac{y_1+y_2+y_3}{3}\\
\bar{z}&=&\frac{z_1+z_2+z_3}{3}\\
\end{eqnarray}
Then the equation can be written in the form
\begin{eqnarray}
(x-\bar{x})^2+(y-\bar{y})^2+(z-\bar{z})^2=R^2
\end{eqnarray}
where $R$ is a constant defined in terms of $r$ and the coordinates of $A,\,B$ and $C$.
This is the equation of a sphere.
Note that OP has amended the question to ask what conditions on $A,\,B$ and $C$ guarantee that the solution will be a sphere.
First let
$$D=\left(\bar{x},\bar{y},\bar{z}\right)$$
The equation can also be written in the form
\begin{equation}
x^2-2x\bar{x}+y^2-2y\bar{y}+z^2-2z\bar{z}+\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)=\frac{1}{3}r^2
\end{equation}
which can be written
\begin{eqnarray}
(x-\bar{x})^2+(y-\bar{x})^2+(y-\bar{y})^2&=&\frac{1}{3}r^2+\bar{x}^2+\bar{y}^2+\bar{z}^2-\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)\\
&=&\frac{1}{3}r^2+\vert D\vert^2 -\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)\\
&>&0
\end{eqnarray}
So the condition is
$$ \vert A\vert^2+\vert B\vert^2+\vert C\vert^2-3\vert D\vert^2<r^2 $$
Thus, if we have $A(2,2,2),\,B(3,3,3),\,C(4,4,4)$ and $r^2=30$, then we have $D(3,3,3)$. So
\begin{eqnarray}
\vert A\vert^2+\vert B\vert^2+\vert C\vert^2-3\vert D\vert^2&=&12+27+48-3(27)\\
&=&6\\
&<&30
\end{eqnarray}
so there will be a sphere for that collection of three points.
Using the previous result
\begin{equation}
R^2=\frac{1}{3}r^2+\vert D\vert^2 -\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)=8
\end{equation}
we get
$$ (x-3)^2+(y-3)^2+(z-3)^2=8 $$
