Question 28
Three sides $OAB$, $OAC$ and $OBC$ of a tetrahedron $OABC$ are right-angled triangles, i.e. $\angle AOB=\angle AOC=\angle BOC=90^\circ$. Given $OA=7$, $OB=2$ and $OC=6$, find the value of
$$[OAB]^2+[OAC]^2+[OBC]^2+[ABC]^2,$$
where $[XYZ]$ denotes the area of triangle $XYZ$.
$$[OAB]^2+[OAC]^2+[OBC]^2+[ABC]^2,$$
where $[XYZ]$ denotes the area of triangle $XYZ$.