Distance between two lines $L_1:\> x+y+z=6,\> x-2z=-5$; $L_2:\> x+2y=3,\> y+2z=3 $

Parametrize the two lines as follows:

For $\mathbb L_{1}$, let $z=t$. Then, $x= -5+2t$ and $y = 11-3t$, which leads to $P_1=(-5,11,0)$ and $v_1=(2,-3,1)$.

For $\mathbb L_{2}$, let $z=s$. Then, $x= 3-2s$ and $y = -3+4s$, which leads to $P_2=(-3,3,0)$ and $v_2=(4,-2,1)$.

Thus, $ v_{1}\times v_{2} = (-1,2,8) $ and the distance is

$$d(\mathbb L_{1},\mathbb L_{2})=\frac{\left|\left(P_{2}-P_{1}\right)\cdot\left(v_{1}\times v_{2}\right)\right|}{\left|v_{1}\times v_{2}\right|} = \frac{\left|(2,-8,0)\cdot(-1,2,8)\right|}{\left|(-1,2,8)\right|} = \frac{18}{\sqrt{69}} $$