Some ideas: for (1) write the lines vector-parametrically equating all the fractions to a real parameter $\,t\,$: $L_1:x=1+2t\,\,,\,\,y=-3+t\,\,,\,\,z=-t\Longrightarrow L_1: (1,-3,0)+t(2,1,-1) $
$L_2: x=-2-2t\,\,,\,\,y=-5+3t\,\,,\,\,z=1-5t\Longrightarrow L_2:\,(-2,-5,1)+t(-2,3,-5)$
Take now any point in $\,L_1\,$ , say $\,(1,-3,0)\,$ and find the point in $\,w\in L_2\,$ for which $\,\overrightarrow{(1,-3,0)w}\perp(2,1,-1)\,$: $w\in L_2\Longrightarrow w=(-2t-2\,,\,3t-5\,,\,-5t+1)\Longrightarrow \overrightarrow{(1,-3,0)w}=(-2t-3\,,\,3t-2\,,\,-5t+1)$ and then we get $(-2t-3\,,\,3t-2\,,\,-5t+1)\perp(2,1-1)\Longrightarrow 0=-4t-6+3t-2+5t-1\Longrightarrow $ $\Longrightarrow 4t=9\Longrightarrow t=\frac{9}{4}$ Now substitute and calculate with the usual formular the distance between the two point. Disclaimer: The above is a lengthy process and there are pre-calcualted formulae. The above is just one way to do it. Another one uses differential calculus, etc.
For two: you need a plane with a normal $\,(a,b,c)\,$ perpendicular to both given planes' normals , i.e. s.t. $(i)\,\,(a,b,c)\cdot (3,-2,1)=0\Longrightarrow\,\,I\,\, 3a-2b+c=0$ $(ii)\,\,(a,b,c)\cdot(5,-4,3)=0\Longrightarrow \,\,II\,\,5a-4b+3c=0$
Now solve this system, say $\,2I-II=a-c=0\Longrightarrow a=c\Longrightarrow \,\,I\,\,\,3a-2b+a=0\Longrightarrow b =2a$
So the wanted plane has as normal any vector of the form $\, (a,2a,a)\,\,,\,\,a\neq 0\,$ , and as we want it to pass throught the point $\,(2,-1,5)\,$ we then get $2a-2a+5a+d =0\Longrightarrow d=-5a$so choosing, say $\,a=1\,$ we get the plane $x+2y+z-5=0$