parmenides51 wrote: $ABCDA'B'CD'$ is a rectangular parallelepiped with $AA'= 2AB = 8a$. $BC$ ?

I added the wording for possible typos in photo

Quote: Knowing that the straight lines $AC'$ and $FD$ are perpendicular, compute the volume of the parallelepiped $ABCDA'B'C'D'$ $AC'\ $ Given the rectangular parallelepiped $ABCDA'B'CD'\ :\ A(t,0,.0),B(t,4a,0),C(0,4a,0),D(0,0,0),A'(t,0,8a),B'(t,4a,8a),C'(0,4a,8a),D'(0,0,8a)$. Midpoint $E(t,2a,0)$, point $M(0,0,a(1+\frac{t}{w})\ )$ with $w=\sqrt{t^{2}+16a^{2}}$. Choose $F(t,0,s)$. Condition $CF+FM$ has a minimum, so $\sqrt{w^{2}+s^{2}}+\sqrt{t^{2}+(a+\frac{at}{w}-s)^{2}}=f(s)$, then $f'(s)=0 \Rightarrow s=a$. Equation of the plane $(DEF)\ :\ 2ax-ty-2tz=0$, equation of the plane $(DB'C')\ :\ -2y+z=0$, $(DEF)\ \bot \ (DB'C')$. Direction vector of $AC'\ :\ (t,-4a,-8a)$, direction vecton of $FD\ :\ (t,0,a)$, $AC' \bot FD \Rightarrow t^{2}-8a^{2}=0 \Rightarrow t=2\sqrt{2}a$. Volume $=64\sqrt{2}a^{3}$.