Vec r=(1,0,-1)+λ(1,2,2) and L2: vec r=(2,3,0)+μ(2,-2,1).

Directions: $\vec d_1=(1,2,2)$, $\vec d_2=(2,-2,1)$, $\vec d_1\cdot\vec d_2=0$, so the lines are perpendicular. $\vec d_1\times\vec d_2=(6,3,-6)=3(2,1,-2)$, $|{\cdot}|=9$. With $\vec{AB}=(1,3,1)$, $\vec{AB}\cdot(6,3,-6)=9$, so the shortest distance is $9/9=1\neq0$: skew lines.
(A) True: distance $1$, and not intersecting.
(B) False: the angle is $90^\circ$, not $\cos^{-1}(4/9)$.
(C) False: the plane through $L_1$ parallel to $L_2$ has normal $(2,1,-2)$ and passes through $(1,0,-1)$: $2x+y-2z=4$, not $2$.
(D) False: solving for the common perpendicular gives $\lambda=1,\ \mu=\frac13$; the feet are $(2,2,1)$ on $L_1$ and $\left(\frac83,\frac73,\frac13\right)$ on $L_2$, so the stated $z$-coordinate $\frac23$ is wrong.