Answer: 163

Solution:

Place $D = (0, 0, 0),$ $C = (12, 0, 0),$ $B = (12, 12, 0),$ $A = (0, 12, 0),$ and $E = (x, y, 12),$ using the given distance from $E$ to face $ABCD.$ From $CE = DE$ we get $x = 6,$ and then $DE = 14$ gives $36 + y^2 + 144 = 196,$ so $y = 4$ and $E = (6, 4, 12).$

In trapezoid $ABFG,$ $\overline{GF}$ is parallel to $\overline{AB}$ with $GF = 6$ and $AG = BF,$ so $G$ and $F$ are symmetric about the plane $x = 6:$ $G = (3, y_2, z_2)$ and $F = (9, y_2, z_2).$ Face $ADEG$ is planar, and the plane through $A,$ $D,$ $E$ contains the entire $y$-axis direction (both $A$ and $D$ have $x = z = 0$), so it is the plane $z = 2x,$ which indeed contains $E.$ Hence $z_2 = 6.$ Now $AG = 8$ gives $3^2 + (y_2 - 12)^2 + 6^2 = 64,$ so $y_2 = 12 \pm \sqrt{19}.$

Then $$\begin{aligned} EG^2 &= 3^2 + (y_2 - 4)^2 + 6^2 \\ &= 45 + \left(8 \pm \sqrt{19}\right)^2 \\ &= 128 \pm 16\sqrt{19}, \end{aligned}$$ and the stated form $p - q\sqrt{r}$ corresponds to $128 - 16\sqrt{19}.$ Thus $p + q + r$ $= 128 + 16 + 19$ $= 163.$