Cutting an Octahedron

Let the points of intersection of the plane with the edges of the octahedron be $A,B,C,D$ . Take the single cut-off vertex of the octahedron as the origin and align it so that its edges lie in the $xz$ and $yz$ planes, on the positive $z$ side.

We can assign coordinates to the points as $A(a,0,a)$ , $B(0,b,b)$ , $C(-c,0,c)$ and $D(0,-d,d)$ , where $a,b,c,d$ are positive real numbers.

Using Pythagoras to work out the side-lengths of the cross-section, we get the following three equations:

$\begin{aligned} a^2-ab+b^2&=\frac{169}{2}\\ b^2-bc+c^2&=56\\ c^2-cd+d^2&=632 \end{aligned}$

Also,

$d^2-da+a^2=\frac{DA^2}{2}$

We want to find the length $DA$ . But we need one more equation; we haven't yet used the fact that these points are coplanar. The scalar triple product of three coplanar vectors is zero; using this condition on the vectors $\overrightarrow{AB}$ , $\overrightarrow{AC}$ and $\overrightarrow{AD}$ , and cancelling down terms, leads to the fourth equation $a b c + a c d = a b d+ b c d$ .

We can now solve these four equations numerically (I used Wolfram|Alpha - there may be a neater way, though) to find $a=\frac{15}{2}\sqrt2$ , $b=4\sqrt2$ , $c=6\sqrt2$ and $d=20\sqrt2$ . Substituting these values in, we find the remaining side $DA=\boxed{35}$ .