Line L1 is given by the planes 2x-y=0 and y-z=0. Line L2 is given by the planes x-2y=3 and 2y+z=0. Let d be the shortest distance between L1 and L2.…

Direction of $L_1$ is the cross product of the plane normals $(2,-1,0)\times(0,1,-1)=(1,2,2)$, and $L_1$ passes through the origin. For $L_2$ the normals $(1,-2,0)$ and $(0,2,1)$ give direction $(-2,-1,2)\parallel(2,1,-2)$, and $(3,0,0)$ lies on it since $3-0=3$ and $0+0=0$. The common-perpendicular direction is $(1,2,2)\times(2,1,-2)=(-6,6,-3)$, of length $9$. With $\vec v=(3,0,0)$ joining the two points, $d=\dfrac{|\vec v\cdot(-6,6,-3)|}{9}=\dfrac{18}{9}=2$. Hence $d^2=4$. The lines are also perpendicular since $(1,2,2)\cdot(2,1,-2)=0$, a useful consistency check. Alternatively one can parametrise both lines, minimise the squared distance between a point on each over the two parameters, and obtain the same minimum value of 4, which verifies the vector formula for the shortest distance between skew lines.