In the [tex]x,y[/tex] plane, we have [tex]z=0[/tex] everywhere. So in the equation of the sphere, we have
[tex]25=(x-2)^2+(y+10)^2+(-3)^2\implies(x-2)^2+(y+10)^2=16=4^2[/tex]
which is a circle centered at (2, -10, 0) of radius 4.
In the [tex]x,z[/tex] plane, we have [tex]y=0[/tex], which gives
[tex]25=(x-2)^2+10^2+(z-3)^2\implies(x-2)^2+(z-3)^2=-75[/tex]
But any squared real quantity is positive, so there is no intersection between the sphere and this plane.
In the [tex]y,z[/tex] plane, [tex]x=0[/tex], so
[tex]25=(-2)^2+(y+10)^2+(z-3)^2\implies(y+10)^2+(z-3)^2=21=(\sqrt{21})^2[/tex]
which is a circle centered at (0, -10, 3) of radius [tex]\sqrt{21}[/tex].
