The sphere S: x^2+y^2+z^2-2x+4y-6z-2=0 is cut by the plane Pi: 2x-y+2z=1 in a circle Gamma.

Completing squares gives $(x-1)^{2}+(y+2)^{2}+(z-3)^{2}=16$, so the centre of $S$ is $C(1,-2,3)$ and its radius is $4$. For $\Pi$ with normal $\vec n=(2,-1,2)$, $|\vec n|=3$, the distance from $C$ is $\frac{|2+2+6-1|}{3}=3$. Hence the radius of $\Gamma$ is $\sqrt{16-9}=\sqrt7$ (A true). The centre of $\Gamma$ is the foot of the perpendicular from $C$: $C-\frac{9}{9}\vec n=(-1,-1,1)$ (B true). Planes parallel to $\Pi$ have the form $2x-y+2z=k$ and touch $S$ when $\frac{|10-k|}{3}=4$, i.e. $k=22$ or $k=-2$; $k=-12$ gives distance $\frac{22}{3}>4$, so it does not touch (C false). The farthest point of $S$ from $\Pi$ lies on the far side of the centre at distance $3+4=7$, not $8$ (D false). Numerical checks agree. Answer: A, B.