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. –
On
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}
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To avoid forgetting solutions it is best to keep the whole system together.
$\begin{cases} (x-2)^2+y^2+z^2=9\\ x^2+(y-2)^2+z^2=9\\ x^2+y^2+(z-2)^2=9\\ \end{cases}\iff\begin{cases} (-4x+4)+x^2+y^2+z^2=9\\ (-4y+4)+x^2+y^2+z^2=9\\ (-4z+4)+x^2+y^2+z^2=9\\ \end{cases}$
Now replacing row $(2)$ by $(2)-(1)$ you get $4x=4y$ and replacing row $(3)$ by $(3)-(1)$ you get $4x=4z$.
But you still have to carry equation $(1)$ for the system to stay complete. This is probably what you forget to do in your solution.
Or maybe you also replaced $(1)$ by a combination of rows, and lost information. In Gauss elimination at least one row should stay untouched at each step.
$\begin{cases} (-4x+4)+x^2+y^2+z^2=9\\ x=y\\ y=z\\ \end{cases} \iff \begin{cases} (-4x+4)+3x^2=9\\ x=y\\ y=z\\ \end{cases}\iff \begin{cases} 3x^2-4x-5=0\\ x=y\\ y=z\\ \end{cases}$
Which solves to $\displaystyle x=y=z=\frac{2\pm\sqrt{19}}3$.
Yes, you are right: $x=y=z$.
Now, $$(x-2)^2+2x^2=9$$ or $$3x^2-4x-5=0,$$ which gives these points.