Let S be the sphere x^2+y^2+z^2-2x-4y-6z+10=0 and let Pi be the plane x+y+z=3. The plane Pi cuts S in a circle C.

The sphere has centre $(1,2,3)$ and radius $\sqrt{1+4+9-10}=2$. The distance of the centre from $\Pi$ is $|6-3|/\sqrt3=\sqrt3$, so the circle has radius $\sqrt{4-3}=1$: (A) true. The centre of $C$ is the foot of the perpendicular, $(1,2,3)-\frac{3}{3}(1,1,1)=(0,1,2)$, not $(1,1,1)$: (B) false. Spheres through $C$ have the form $S+\lambda(x+y+z-3)=0$; passing through the origin needs $10-3\lambda=0$, so $\lambda=\frac{10}{3}$. Its centre is $\left(-\frac23,\frac13,\frac43\right)$ and its radius squared is $\frac{1+4+16}{9}-0=\frac{21}{9}$, so the radius is $\frac{\sqrt{21}}{3}$: (C) true. Since $\Pi$ cuts $S$, points of $S$ lie on $\Pi$ and the least distance is $0$, not $2-\sqrt3$: (D) false.