Finding the radius of a sphere inscribed in a right prism 
We have right prism $ABCA_{1}B_{1}C_{1}$ and points $E$, $D$ such that: 
  
  
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*$A_{1}E:EB_{1}=B_{1}D:DC_{1}=1:2$ 
  
*The distance between lines $AE$ and $BD$ is $\sqrt{13}$. 
  
  
  Find the radius of the sphere inscribed in this prism.

My attempt: I made a translation of$AE$ through the $AA_{1}B_{1}B$ while $AE$ doesn't intersect $BD$. Let's call new line madden by $AE$ : $BE_{1}$. So I get a new area made by translation : $BDE_{1}$. 
So $AE$ parallel to this area then using the distance between two lines I determined the length of triangle(in base) : $a = 2\sqrt{13}$.
Then using projection of sphere on my base-triangle I found the radius $=\sqrt{\frac{13}{2}}$. 
But the answer is $\frac{13}{6}$. Please find a mistake in solution ?
 A: The distance between lines $AE$ and $BD$ is the same as the distance of point $A$ from plane $BDE_1$, which turns out to be $\sqrt{3/13}a$, where $a$ is the side length of the base. 
To see why, observe first of all that line $DE_1$ intersects $A_1C_1$ at a point $H_1$ such that $C_1H_1=a/3$. The line through $B$ parallel to $DE_1$ intersects then $AC$ at a point $H$ such that $CH=a/2$. Line $BH$ is the intersection between planes $BDE_1$ and $ABC$ and is perpendicular both to $AC$ and to $HH_1$. The perpendicular projection of $A$ onto plane $BDE_1$ must then belong to line $HH_1$.
Consider now point $A_2$ on $A_1C_1$ such that $A_2H_1=a/2$: $AHH_1A_2$ is a parallelogram by construction and we need to find its altitude $AK$ with respect to base $HH_1$. But that is easy, once you notice that 
$AH\cdot AA_1=AK\cdot HH_1$ and that $HH_1=AA_2=(\sqrt{13}/6)a$.
We then obtain $AK=\sqrt{3/13}a$ and equating that to $\sqrt{13}$ one gets $a=13/\sqrt3$, so that in the end: $r =(\sqrt3/6) a=13/6$.
