How far to a pyramid

If the coordinates of $P$ are $(x,y,z)$ and the numbers $3,7,9,11$ are $d_1,d_2,d_3,d_4$ in some order, then $\begin{array}{rclcrcl} (x-\tfrac12s)^2 + (y - \tfrac12s)^2 + z^2 & = & d_1^2 & \hspace{2cm}& (x + \tfrac12s)^2 + (y - \tfrac12s)^2 + z^2 & = & d_2^2 \\ (x+\tfrac12s)^2 + (y + \tfrac12s)^2 + z^2 & = & d_3^2 & & (x - \tfrac12s)^2 + (y + \tfrac12s)^2 + z^2 & = & d_4^2 \end{array}$ Subtracting these equations in pairs, we see that

$2xs \; = \; d_2^2 - d_1^2 \; = \; d_3^2 - d_4^2 \hspace{2cm} 2ys \; = \; d_4^2 - d_1^2 \; = \; d_3^2 - d_2^2$

and this can only happen if $d_1^2 + d_3^2 = d_2^2 + d_4^2$ , so we have $d_1=3,d_2=7,d_3=11,d_4=9$ , without loss of generality. Thus $x = 20s^{-1}$ and $y = 36s^{-1}$ , and so we have $\begin{array}{rcl} (20s^{-1} - \tfrac12s)^2 + (36s^{-1} - \tfrac12s)^2 + z^2 & = & 9 \\ 1696s^{-2} + \tfrac12s^2 + z^2 & = & 65 \end{array}$ while the distance $e$ from $P$ to the vertex $(0,0,\tfrac{1}{\sqrt{2}}s)$ satisfes $\begin{array}{rcl} e^2 & = & x^2 + y^2 + (z - \tfrac{1}{\sqrt{2}}s)^2 \; = \; x^2 + y^2 + z^2 + \tfrac12s^2 - sz\sqrt{2} \\ & = & 1696s^{-2} + \tfrac12s^2 + z^2 - sz\sqrt{2} \; = \; 65 - sz\sqrt{2} \end{array}$ Now $\begin{array}{rcl} 3392 + s^4 + 2s^2z^2 & = & 130s^2 \\ (s^2 - 65)^2 + 2s^2z^2 & = & 833 \\ 2s^2z^2 & = & 833 - (s^2 - 65)^2 \; \le \; 833 \end{array}$ so that $|sz\sqrt{2}| \le \sqrt{833} = 7\sqrt{17}$ , and hence $\sqrt{65 - 7\sqrt{17}} \; \le \; e \; \le \; \sqrt{65 + 7\sqrt{17}}$ making the answer $65 + 7 + 17 = \boxed{89}$ .

Let $A(0,\frac{s}{\sqrt{2}},0), B(\frac{s}{\sqrt{2}},0,0),C(0,-\frac{s}{\sqrt{2}},0),D(-\frac{s}{\sqrt{2}},0,0)$ be vertices of the pyramid's base, $E(0,0,\frac{s}{\sqrt{2}})$ the pyramid's apex, $P(x,y,z)$ the point in space, $AP=7, BP=3, CP=9, DP=11$ , $e=EP$ .

Equations for the distances $AP,BP,CP,DP,EP$ can be manipulated to get a single equation

$64 (-53 x⁴ + 3185x² - 38416) - (x² e² - 65x²)²=0$

Ignoring negative values of $e$ we find that minimum value of $e$ is $\sqrt{65-7\sqrt{17}}$ and maximum is $\sqrt{65+7\sqrt{17}}$ .

$a=65, b=7, c=17$ . The rounded values are $6.0115$ for minimum and $9.6882$ for maximum.