Find all the points that lie $ 3$ units from each of the points $ (2,0,0), (0,2,0), \text{ and }(0,0,2)$ 
I calculated the result $$ \{(x,y,z) \in\mathbb{R^3}: x=y=z\}.$$
I'm wondering whether I did this problem correctly and if I did how to draw the set of solutions. 
I used the euclidean distance formula with the square root (the euclidean norm) and set up three equations for each point $(2,0,0)$, $(0,2,0)$ and $(0,0,2)$ that made sure the distance from arbitrary $(x,y,z)$ to these three points equal to  $3$ and arrived at the solution set. –

The set of all points equidistqnt from $A=(2,0,0)$ and $B=(0,2,0)$:
\begin{align}
   \sqrt{(x-2)^2+y^2+z^2}&=\sqrt{x^2+(y-2)^2+z^2} \\
   -4x+4 &= -4y+4 \\
   x&=y
\end{align}
Similarly, the set of all points equidistant from $A=(2,0,0)$ and $C=(0,0,2)$ is described by $x=z$.
Hence the set of all points equidistant from $A, B$, and $C$ is described by $x=y=z$, that is, the line P(t) = (t,t,t). If that distance is $3$, then $d(A,P(t))=3$ implies
\begin{align}
   \sqrt{(t-2)^2+t^2+t^2}&=3 \\
   3t^2-4t+4 &= 9 \\
   3t^2-4t-5 &= 0 \\
   t &= \frac{2 \pm \sqrt{19}}{3}
\end{align}
