Vec r=λ(hat i+2hat j+2hat k) and L2: vec r=(3hat i+3hat k)+μ(2hat i+hat j-2hat k), where λ,μ∈mathbb R.

Directions are $\vec d_1=(1,2,2)$ and $\vec d_2=(2,1,-2)$. Since $\vec d_1\cdot\vec d_2=2+2-4=0$ the lines are perpendicular. The common normal direction is $\vec d_1\times\vec d_2=(-6,6,-3)\parallel(2,-2,1)$ with magnitude $9$. The vector from $(0,0,0)$ to $(3,0,3)$ has dot product $-27$ with $(-6,6,-3)$, so the shortest distance is $27/9=3\ne0$; lines are skew, hence A and B are true. The plane through the origin with normal $(2,-2,1)$ contains $L_1$ ($2-4+2=0$) and is parallel to $L_2$ ($4-2-2=0$), so D is true. Solving the perpendicularity conditions gives $\lambda=1,\mu=0$: the feet are $(1,2,2)$ and $(3,0,3)$, so C, which claims $(2,4,4)$, is false.