By a simple variant of the British Flag Theorem, if $A B C D$ is a square and $P$ any point in space, $A P^{2}+C P^{2}=B P^{2}+D P^{2}$. Four of the five given vertices must form a square $A B C D$, and by experimentation we find their distances to the given point $P$ must be $A P=2, B P=6, C P=9, D P=7$. Then $A, C$, and the other two vertices $E, F$ also form a square $A E C F$, so $85=A P^{2}+C P^{2}=E P^{2}+F P^{2}=8^{2}+F P^{2} \Rightarrow F P=\sqrt{21}$.