The line L1: (x-1)/(1)=(y-2)/(-1)=(z-3)/(2) and the line L2 passing through (0,1,-1) with direction ratios 2,1,-1 are skew. Let Pi be the plane that…

The normal of $\Pi$ is perpendicular to both direction vectors, so $\vec n=(1,-1,2)\times(2,1,-1)=(-1,5,3)$ and $|\vec n|=\sqrt{35}$. Since $\Pi$ contains the point $(1,2,3)$ of $L_1$, its equation is $-(x-1)+5(y-2)+3(z-3)=0$, i.e. $-x+5y+3z-18=0$. For the point $(2,0,1)$: $-2+0+3-18=-17$, so the distance is $\frac{17}{\sqrt{35}}$. Note that the shortest distance between $L_1$ and $L_2$ (the distance of $(0,1,-1)$ from $\Pi$) is $\frac{16}{\sqrt{35}}$, which is option (A); it answers a different question. Options (B) and (D) come from an arithmetic slip in the normal's length or in the constant term.