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

Let $\vec d_1=(1,1,2)$, $\vec d_2=(2,-1,1)$. Then $\vec d_1\times\vec d_2=(3,3,-3)=3(1,1,-1)$. The vector joining the base points is $(1,-2,-2)$, whose dot product with $(1,1,-1)$ is $1$. Since it is nonzero, the lines are skew and (A) is false; the distance is $\frac{|1|}{\sqrt3}=\frac1{\sqrt3}$, so (B) is true. The plane through $(1,2,3)$ with normal $(1,1,-1)$ is $x+y-z=0$ (and it contains $L_1$ because $\vec d_1\cdot(1,1,-1)=0$), so (C) is true. Solving $(P-Q)\cdot\vec d_1=(P-Q)\cdot\vec d_2=0$ gives $t=-\frac43$, $s=-1$, i.e. the foot on $L_1$ is $\left(-\frac13,\frac23,\frac13\right)$ and on $L_2$ is $(0,1,0)$, so (D) is true. Answer: B, C, D.