how do i get the position of the craft relative to the universe?

Oh, you just want distances.

To do that, you need to add your position to the solar position of your celestial body; this is value x. Value y is the target plus their celestial body’s solar position. You will find the dist function in the vector operations function. Slap in x to one side and y to the other, and this should yield absolute distance.

Best bet is using the central body and trigonometry.
For instance, a coordinate of (84,-160,35000) will translate to (-1251,-3437,34808) if we assume that a coordinate above (0,0,x) represents the direction of the x axis and (90,x,y) represents the z axis. Note that this original vector is the PCI coordinates of the player craft added to its celestial body's solar position.

The z value is taken from the sin of the latitude angle multiplied with the PCI vector magnitude. The x value is taken from the cosine of the latitude multiplied by the result of the absolute value of the cosine of latitude multiplied by the magnitude (we’ll just call this w), and y is taken from the sine of the latitude multiplied with w.

O.k. I don't know "toblox lua" sorry, only Vizzy. There you can subtract 3D-vectors simply by using the " - " operator.

@TomKerbal
in roblox lua, the distance between two vectors is:
(Vector3.new(x, y, z) - Vector3.new(x, y, z).Magnitude

Nav/Position and Nav/TargetPosition given in PCI coordinates.
Coordinates of Center of Mass (CoM) wrt Celestial Body center (which is the same as CoG, Center of Gravity here). PCI is Planet-Centered Inertial Frame of Reference.
Trivially the magnitude of the vector difference is the distance between them.