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

Direction vectors $\vec d_1=(2,3,1)$, $\vec d_2=(1,-1,2)$; $\vec d_1\times\vec d_2=(7,-3,-5)$, magnitude $\sqrt{83}$. Points $(1,-1,2)\in L_1$ and $(0,1,-2)\in L_2$; their difference is $(-1,2,-4)$ with dot product $-7-6+20=7$ with the normal. So the shortest distance is $7/\sqrt{83}\ne0$, lines are skew. (A) False: the distance is non-zero. (B) True. (C) False: $\sqrt{59}$ is wrong magnitude. (D) False: the plane through $(1,-1,2)$ with normal $(7,-3,-5)$ is $7x-3y-5z=7+3-10=0$, i.e. $7x-3y-5z=0$; the constant $7$ belongs to the plane containing $L_2$.