jmc

$PQ=\sqrt{(8-7{)}^2+(8-12{)}^2+(1-10{)}^2}=\sqrt{98}$ $PR=\sqrt{(11-7{)}^2+(3-12{)}^2+(9-10{)}^2}=\sqrt{98}$ $QR=\sqrt{(11-8{)}^2+(3-8{)}^2+(9-1{)}^2}=\sqrt{98}$ 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 $6a^2=\boxed{294}$.