It sounds like you basically have to solve a multi-equation system. Since all the lines are straight lines, you are solving a linear algebra problem. If you are working in 3D space (z,y,z) then it means that the equation of a line is r(t) = at + b(1-t) where a and b are points on the line. You have one of the two points for each line (the base point) and you also know that the second point is the same for all lines (since they all meet at that point).
So basically you have to solve for that last point that is common to all lines but which has a different value of t (for that particular linear equation). You can however get the value of t by using the distance. If you have r(t) = a + (b-a)*t when t = distance from base (which you have). Now given the above you have three equations in three unknowns and you can solve for the missing point.
In summary you have:
r(t)_i = a_i + (b_i-a_i)*d_i for the ith line and you have a_i and d_i and b_i = b for all i.