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.

To express that the center is equidistant from two points, you write
$$(x_c-x_0)^2+(y_c-y_0)^2+(z_c-z_0)^2=(x_c-x_1)^2+(y_c-y_1)^2+(z_c-z_1)^2.$$
This can be rewritten as
$$(2x_c-x_0-x_1)(x_1-x_0)+(2y_c-y_0-y_1)(y_1-y_0)+(2z_c-z_0-y_1)(z_1-z_0)=0,$$ which is linear in $x_c,y_c,z_c$. This is in fact the equation of the bissector plane of the two points.
By considering the equations of two bissectors and that of the plane through the given points, you get a $3\times3$ linear system.

For efficient computation, you can translate all points so that $P_0$ comes to the origin. Then
$$\left(\begin{matrix}x'_1&y'_1&z'_1\\x'_2&y'_2&z'_2\\y'_1z'_2-y'_2z'_1&z'_1x'_2-z'_2x'_1&x'_1y'_2-x'_2y'_1\end{matrix}\right)
\left(\begin{matrix}2x'_c\\2y'_c\\2z'_c\end{matrix}\right)=
\left(\begin{matrix}{x_1'}^2+{y_1'}^2+{z_1'}^2\\{x_2'}^2+{y_2'}^2+{z_2'}^2\\0\end{matrix}\right).$$
