Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$ Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$

Let the circle passes through the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ and the plane $Ax+By+Cz+D=0$
So the equation of the circle is $x^2+y^2+z^2+2ux+2vy+2wz+d+\lambda(Ax+By+Cz+D)=0$
As it passes through $(-1,0,0)$ so
$1-2u+d+\lambda(-A+D)=0$
As it passes through $(0,2,0)$ so
$4+4v+d+\lambda(2B+D)=0$
As it passes through $(0,0,3)$ so
$9+6w+d+\lambda(3C+D)=0$
Here i am stuck.I cannot find the radius of the circle.Is my method correct or not?Is there simpler method possible?Please help.

Let $\Omega(x_0,y_0,z_0)$ be the center of this circle, and let us denote our points by $A,B$ and $C$, respectively. Then, by definition, we have the following family of equalities: $$\begin{cases}\operatorname{dist}(\Omega,A)=\operatorname{dist}(\Omega,B)\\
\operatorname{dist}(\Omega,C)=\operatorname{dist}(\Omega,B)\\
\operatorname{dist}(\Omega,C)=\operatorname{dist}(\Omega,A)
\end{cases}\iff\begin{cases}(x_0+1)^2+y_0^2+z_0^2=x_0^2+(y_0-2)^2+z_0^2\\
x_0^2+y_0^2+(z_0-3)^2=x_0^2+(y_0-2)^2+z_0^2\\
x_0^2+y_0^2+(z_0-3)^2=(x_0+1)^2+y_0^2+z_0^2
\end{cases}$$
If we add the condition that $\Omega$ is in the plane formed by these three points, then this should be sufficient to determine the coordinates of the center of the circle.
