Vec r=(hat i+2hat j+3hat k)+t(hat i-hat j+2hat k) and L2: vec r=(2hat i+hat k)+s(2hat i+hat j-hat k). Let Pi be the plane containing L1 and parallel…

A normal to $\Pi$ must be perpendicular to both direction vectors, so $\vec n=(\hat i-\hat j+2\hat k)\times(2\hat i+\hat j-\hat k)$. Expanding the determinant gives $\vec n=(1-2)\hat i-(-1-4)\hat j+(1+2)\hat k=-\hat i+5\hat j+3\hat k$, with $|\vec n|=\sqrt{35}$. The plane passes through $A(1,2,3)$ on $L_1$, so its equation is $-(x-1)+5(y-2)+3(z-3)=0$. The vector $\overrightarrow{AP}=(3,-1,-5)$ gives $\vec n\cdot\overrightarrow{AP}=-3-5-15=-23$. Hence the distance is $\dfrac{23}{\sqrt{35}}$. The value $\dfrac{17}{\sqrt{35}}$ is the distance of the point $(2,0,1)$ of $L_2$ from $\Pi$, which equals the shortest distance between the two skew lines, so it answers a different question. The other choices come from an arithmetic slip in $\vec n$.