Let L1: (x-1)/(1)=(y-2)/(2)=(z-3)/(2) and L2: (x-3)/(2)=(y-1)/(1)=(z)/(-2).

Directions are $\vec d_1=(1,2,2)$ and $\vec d_2=(2,1,-2)$. Their dot product is $2+2-4=0$ (so the lines are perpendicular, not at $60^\circ$; C false) and $\vec d_1\times\vec d_2=(-6,6,-3)=-3(2,-2,1)$. Take $\vec n=(2,-2,1)$, $|\vec n|=3$, and $\overrightarrow{AB}=(2,-1,-3)$ between the points $A(1,2,3)\in L_1$, $B(3,1,0)\in L_2$. Then $\overrightarrow{AB}\cdot\vec n=4+2-3=3$, so the shortest distance is $\frac33=1\neq0$: the lines are skew (A true) and the distance is $1$ (B true). The plane through $A$ with normal $\vec n$ contains $L_1$ and is parallel to $L_2$: $2x-2y+z=2-4+3=1$ (D true).