I'm not sure what the question is, but I can make a few suggestions.
r1r2+r3r2+r3r1=s2 could be a box with dimensions r1×r2×r3. The surface area of the box is 2(r1r2+r3r2+r3r1)=2s2. If all the dimensions and s are whole numbers, then we can find out what the dimensions of the box could be.
r3=(s2-r1r2)/(r1+r2). If r1r2=n2, that is r1r2 is a perfect square, then (s-n)(s+n)/(r1+r2). We can make r1+r2=s-n or s+n, so that r3=s+n or s-n respectively.
The perfect squares are 1, 4, 9, 16, 25, etc. All we need is to find r1r2 which happens to be a perfect square. For example, 1×16=2×8=4×4=16=42.
Let's use 1 and 16, then n=4. The sum of 16 and 1 is 17 so s-4=17 or s+4=17. Therefore, s=21 or 13. r3 is respectively 25 or 17.
Let's check: r1=1, r2=16, s=21, r3=25. r1r2+r3r2+r3r1=16+400+25=441=212=s2. The dimensions of the box are 1×16×25.
We can pick another set based on r1=2, r2=8, so r1r2=16, r1+r2=10; n=4, s+4=10, s=6; r3=2.
Check: 2×8+2×8+4=62. Box dimensions are 2×8×2.
In practical terms then we could start with two square pieces of cardboard (each side s) and make a box with integer dimensions out of the cardboard by following the relationships illustrated above.
A variation on this is to make a box seat (dimensions r1×r2×r3) in the corner of a room, so that it only needs two sides and a top made out of a square piece (side s) of wood. Only three surfaces of the box seat are exposed (for painting, for example).