Almost A Square, But Not Quite

I put coordinates on everything: wlog $A = (0,0,0)$ and the planes are $x = 0$ and $y = 0$ . If $B = (b_1, b_2, b_3)$ , then $b_2^2 + b_3^2 = 123$ , $b_1^2 + b_3^2 = 123$ , and $b_1^2 + b_2^2 + b_3^2 = 144$ . So $b_1^2 = b_2^2 = 21$ and $b_3^2 = 102$ . After some reflections, we can assume that $B = (\sqrt{21}, \sqrt{21}, \sqrt{102})$ .

Proceeding similarly (and using reflections to choose signs wlog), we can get $D = (-\sqrt{102},-\sqrt{102},\sqrt{21})$ and $C = ( \sqrt{21}-\sqrt{102},\sqrt{21}-\sqrt{102},\sqrt{21}+\sqrt{102})$ . Then $AB = CD = 12$ and $BC=AD=15$ , so the sum is $\fbox{54}$ .

I'm sure there is a much nicer solution...?

first AB=(123)^0.5 ; AD=AB/ SIN 45 = (246)^0.5 ; when AB = 12 it is >(123)^0.5 ; so let denote new AB WITH A1B1 ; in traingle A A1 D ; AA1=[ 12-(123)^0.5 ]/2 ; A1D = [AD^2 + AA1^2 ]^0.5 =15.69 ; SO perimeter = 12+ 2* 15.69 +(123)^0.5 =54.47

Imgur I too got the same answer first (approx 54), by dividing the projected area = 123 sqrt(2) by 12. But then thought it may not be correct for the following -

To get the side AB = 12 for ABCD (which has to be a parallelogram), the projections A1B1 and A2B2 of AB on the two planes, cannot be parallel to the line of intersection (the 'fold line') of the two planes. Because that would make ABCD a rectangle with two of the sides equal to sqrt(123) and other two sqrt(2) times that.

So let theta = angle made by one side of the square with the 'fold line'. and

L = side of the squares = sqrt(123).Then $12 = \sqrt{L^2 \cos^2 \theta + L^2 \sin^2 \theta + L^2 \sin^2 \theta}$ $\theta = \sin^-1 \sqrt{\frac{144}{123}-1} = 24.4$ If phi = 90 - theta, then the other sides (AD & BC) of the parallelogram ABCD would be $AD = \sqrt{L^2 \cos^2 \phi + L^2 \sin^2 \phi + L^2 \sin^2 \phi} = 15$ Giving the perimeter = 2(12+15) = 54

So the perimeter is exactly and not approximately 54 as I initially thought.