PARTIAL ANSWER
This how far I've gotten. I hope someone with more a recent education than I can take correct what I have wrong and finish it up.
tl;dr
Here are pieces I think will help:
$$D_A^2=49=r_P^2+z_S^2-2z_Sz_A$$
$$D_B^2=81=r_P^2+z_S^2-2z_Sz_B$$
$$D_C^2=121=r_P^2+z_S^2-2z_Sz_C$$
$$D_D^2=??=r_P^2+z_S^2-2z_Sz_D$$
$$5z_A-9z_B+4z_C=0$$
$$\frac{1}{z_S}=\frac{z_A-z_B}{4}=\frac{z_A-z_C}{9}=\frac{z_B-z_C}{5}$$
$$r_C^2=2r_Ph$$
$$r_P^2=\pm(x_Ax_B+y_Ay_B+z_Az_B)$$
$$r_P^2=\pm(x_Cx_D+y_Cy_D+z_Cz_D)$$
Name the 4 points A, B, C, D and their coordinates as $(x_A,y_A,z_A)$, etc.
Name the ship S and its coordinates are $(x_S,y_S,z_S)$
Name the planet P and its coordinates are $(x_P,y_P,z_P)$ with a radius $r_P$
For convenience, set the center of the planet $P$ to be at the origin.
In addition, orient the axes such that the ship $S$ is directly above the planet along the z-axis.
This tells us that $x_P=y_P=z_P=x_S=y_S=0$
The general equation for the distance between two points in 3D space is
$$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$
Therefore, the distance between the ship $S$ and point $A$ is
$$D_A=7=\sqrt{(x_S-x_A)^2+(y_S-y_A)^2+(z_S-z_A)^2}$$
$$D_A=7=\sqrt{(0-x_A)^2+(0-y_A)^2+(z_S-z_A)^2}$$
$$D_A=7=\sqrt{x_A^2+y_A^2+(z_S^2-2z_Sz_A+z_A^2)}$$
$$D_A=7=\sqrt{(x_A^2+y_A^2+z_A^2)+(z_S^2-2z_Sz_A)}$$
Now, since the planet's center is at the origin, we know $x_A^2+y_A^2+z_A^2=r_P^2$
We can plug that in and square both sides to get
$$D_A^2=49=r_P^2+z_S^2-2z_Sz_A$$
It is trivial to expand that to the other 3 points
$$D_B^2=81=r_P^2+z_S^2-2z_Sz_B$$
$$D_C^2=121=r_P^2+z_S^2-2z_Sz_C$$
$$D_D^2=??=r_P^2+z_S^2-2z_Sz_D$$
Right now, we're sitting at 4 equations and 7 unknowns ($r_P,z_S,z_A,z_B,z_C,z_D,D_D$). That's close but we're not there yet. We can play around with the formulas for $D_A$ and $D_B$ because they're values are known.
$$D_A^2+32 = 49+32=81=D_B^2$$
$$r_P^2+z_S^2-2z_Sz_A+32=r_P^2+z_S^2-2z_Sz_B$$
$$-2z_Sz_A+32=-2z_Sz_B$$
$$2z_Sz_B+32=2z_Sz_A$$
$$z_Sz_B+16=z_Sz_A$$
We can do the same thing for $D_A$ and $D_C$ as well as $D_B$ and $D_C$
$$z_Sz_C+36=z_Sz_A$$
$$z_Sz_C+20=z_Sz_B$$
If we solve all three for $z_S$, we can set them equal.
$$z_S=16/(z_A-z_B)=36/(z_A-z_C)=20/(z_B-z_C)$$
$$z_S=4/(z_A-z_B)=9/(z_A-z_C)=5/(z_B-z_C)$$
$$\frac{1}{z_S}=\frac{z_A-z_B}{4}=\frac{z_A-z_C}{9}=\frac{z_B-z_C}{5}$$
This gives us the relationship
$$5z_A-9z_B+4z_C=0$$
Let's look at the second observation. The formula for the surface area a spherical cap $A_P$ that is distance $h$ from the origin and the area of a circle $A_C$ are given by
$$A_P=2\pi r_Ph$$
$$A_C=\pi r_C^2$$
We're told those are equal so we can drop the $\pi$ to get the equality
$$r_C^2=2r_Ph$$
Unfortunately, this introduces only 1 new equation and 2 new unknowns. I think the solution will involve some trig to find $h$ based on $r_P$, $z_S$, and tangent lines but I don't have the brainpower for that right now.
I'm sure that the probes landing on opposite sides matters but I couldn't get it to do anything useful because it added more unknowns instead of removing them. Here's what I wrote for that section:
We also know that the probes landed on opposite sides of the sphere. That means the line connecting them passes through the center which means it is of length $2r_P$. That gives us the following for the distance between probe $A$ and $B$. (Here I have to assume that the first two distances are for the first two probes. If that's not the case, this section must be reworked.)
$$2r_P=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2+(z_A-z_B)^2}$$
$$2r_P=\sqrt{(x_A^2-2x_Ax_B+x_B^2)+(y_A^2-2y_Ay_B+y_B^2)+(z_A^2-2z_Az_B+z_B^2)}$$
$$2r_P=\sqrt{(x_A^2+y_A^2+z_A^2)+(x_B^2+y_B^2+z_B^2)-2(x_Ax_B+y_Ay_B+z_Az_B)}$$
$$2r_P=\sqrt{(r_P^2)+(r_P^2)-2(x_Ax_B+y_Ay_B+z_Az_B)}$$
$$2r_P=\sqrt{2r_P^2-2(x_Ax_B+y_Ay_B+z_Az_B)}$$
$$4r_P^2=\pm2(r_P^2-(x_Ax_B+y_Ay_B+z_Az_B))$$
$$r_P^2=\pm(x_Ax_B+y_Ay_B+z_Az_B)$$
(I'm not super-confident in how the $\pm$ comes in once you square both sides but I think it has to be considered somewhere.)
We can do the same thing for the other two probes
$$r_P^2=\pm(x_Cx_D+y_Cy_D+z_Cz_D)$$