Put $A$ at the origin with the cube’s edges along the axes and side length $s;$ expand each squared distance in terms of $AP^2,$ $s,$ and the coordinates of $P$

The combination $BP^2 + CP^2 + DP^2 - GP^2$ collapses to $2\,AP^2:$ all terms involving $s$ cancel

Let $A$ be the origin with $B = (s, 0, 0),$ $C = (0, s, 0),$ $D = (0, 0, s),$ $G = (s, s, s),$ and $P = (x, y, z).$ Expanding, $$\begin{aligned} BP^2 &= AP^2 - 2sx + s^2, \\ CP^2 &= AP^2 - 2sy + s^2, \\ DP^2 &= AP^2 - 2sz + s^2, \end{aligned}$$ while $GP^2 = AP^2 - 2s(x + y + z)$ $+ 3s^2.$ Therefore $$\begin{aligned} &BP^2 + CP^2 \\ &\quad {}+ DP^2 - GP^2 \\ &= 2\,AP^2, \end{aligned}$$ with every term involving $s$ or the coordinates of $P$ cancelling.

The given lengths yield $BP^2 = 36000,$ $CP^2 = 18000,$ $DP^2 = 28800,$ and $GP^2 = 9072,$ so $$\begin{aligned} 2\,AP^2 &= 36000 + 18000 \\ &\quad {}+ 28800 - 9072 \\ &= 73728, \end{aligned}$$ giving $AP^2 = 36864$ and $AP = 192.$