Solution:

Compute the squared distances: $PQ^2 = 1^2 + 4^2 + 9^2 = 98,$ $QR^2 = 3^2 + 5^2 + 8^2 = 98,$ and $RP^2 = 4^2 + 9^2 + 1^2 = 98.$ So $P,$ $Q,$ and $R$ form an equilateral triangle with side $\sqrt{98} = 7\sqrt{2}.$

Three mutually equidistant vertices of a cube must be joined by face diagonals, and a face diagonal of a cube with edge $s$ has length $s\sqrt{2}.$ Thus $s = 7,$ and the surface area is $6 \cdot 7^2 = 294.$