(x-1)/(1)=(y-2)/(1)=(z-3)/(2) and L2: x/2=(y-1)/(-1)=(z+1)/(1).

Directions are $\vec d_1=(1,1,2)$ and $\vec d_2=(2,-1,1)$, and $\vec n=\vec d_1\times\vec d_2=(3,3,-3)\parallel(1,1,-1)$. The vector between the points $(1,2,3)$ and $(0,1,-1)$ is $(-1,-1,-4)$; its dot product with $(3,3,-3)$ is $6\ne0$, so the lines are not coplanar and are skew (A true). The shortest distance is $\frac{|6|}{\sqrt{27}}=\frac2{\sqrt3}$ (B true). Solving for the feet of the common perpendicular gives $(-\frac23,\frac13,-\frac13)$ on $L_1$ and $(0,1,-1)$ on $L_2$, not $(2,0,0)$ (C false). The plane through $(0,1,-1)$ with normal $(1,1,-1)$ is $x+y-z=2$ (D true).