AIME 2002 II · 第 2 题

Solution

$PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}$ $PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}$ $QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}$

So, $PQR$ is an equilateral triangle. Let the side of the cube be $a$.

$a\sqrt{2}=\sqrt{98}$

So, $a=7$, and hence the surface area is $6a^2=\boxed{294}$.