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

This answer was written mostly to convince myself that this can be brute-forced by Lagrange multipliers. Anyway, we are trying to minimize $$f(x,y,z,x',y',z') = (x-x')^2+(y-y')^2+(z-z')^2$$ with respect to constraints $$ begin{cases} g_1(x,y,z,x',y',z') = x+y+z-6=0 & \ g_2(x,y,z,x',y',z') = x-2z+5 =0& \ end{cases} qquad begin{cases} g_3(x,y,z,x',y',z') = x'+2y'-3 & \ g_4(x,y,z,x',y',z') = y'+2z'-3 & \ end{cases}$$ The point $(x,y,z,x',y',z')$ which minimizes $f$ will satisfy $$nabla f in operatorname{span}{nabla g_1, nabla g_2,nabla g_3,nabla g_4}.$$ The gradients are $$nabla f (x,y,z,x',y',z') = 2(x-x',y-y',z-z',x'-x,y'-y,z'-z)$$ $$nabla g_1 (x,y,z,x',y',z') = (1,1,1,0,0,0)$$ $$nabla g_2 (x,y,z,x',y',z') = (1,0,-2,0,0,0)$$ $$nabla g_3 (x,y,z,x',y',z') = (0,0,0,1,2,0)$$ $$nabla g_4 (x,y,z,x',y',z') = (0,0,0,0,1,2)$$ so denoting $(mathbf{x}, mathbf{y}, mathbf{z}) = (x-x',y-y',z-z')$ it follows that $$(mathbf{x}, mathbf{y}, mathbf{z}) in operatorname{span}{(1,1,1),(1,0,-2)} cap operatorname{span}{(1,2,0),(0,1,2)}.$$ This intersection can be seen to be $operatorname{span}{(-1,2,8)}$ since $$(-1,2,8) = 2(1,1,1)-3(1,0,-2) = -(1,2,0)+4(0,1,2).$$ Therefore $(mathbf{x}, mathbf{y}, mathbf{z}) = t(-1,2,8)$ for some $t in Bbb{R}$. We also know $$ begin{cases} x+y+z=6 ,mid cdot ,2& \ x-2z=-5 ,mid cdot ,(-3) & \ end{cases} qquad begin{cases} x'+2y'=3 ,mid cdot ,(-1) & \ y'+2z'=3 ,mid cdot ,4 & \ end{cases}$$ so $$-x+2y+8z=27, qquad -x'+2y'+8z'=9.$$ Subtracting gives $$18 = -mathbf{x}+2mathbf{y}-8mathbf{z} = t|(-1,2,8)|^2 = 69t implies t = frac{18}{69}.$$ Now we have $$f(x,y,z,x',y',z') = |(mathbf{x}, mathbf{y}, mathbf{z})|^2 = t^2|(-1,2,8)|^2 = frac{18^2}{69}$$ so finally $$d = frac{18}{sqrt{69}}.$$

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

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

Sketch of a general solution :

1. step one : find $P_1$ and $P_2$ by solving the two systems of equations. In the general case the plane equations have three variables (this is why their intersection is a line meaning an infinity of points) so you need to set for example $z=0$

2. step two : in order to find the directional vectors $v_1$ and $v_2$ we start with the two pairs of normal vectors to the planes whose intersections are the lines we’re studying. So for example with the first line the two normal vectors are $n_1=(1,1,1)^T$ and $n_1’=(1,0,-2)^T$ and a directional vector of the first line is orthogonal to the two normal vectors so for example we can take $v_1=n_1times n_1’$ the vector product of the two normal vectors. Similarly $v_2=n_2times n_2’$

3. step three : With all this in hand we can apply the formula