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Question: interpolation of 3D data
Answer: Mathematica's interpolation function, Interpolation, works on multidimensional data. For example,

data = Flatten[Table[{x, y, x^2 + y^2}, {x, -10, 10}, {y, -10, 10}], 1];
int = Interpolation[data];


Then, you can extract the values for values between the data points:

int[1.1, 1.1]
(* ==&gt; 2.42 *)


And Plot3D, or whatever else you want.

Plot3D[int[x, y], {x, -10, 10}, {y, -10, 10}]




Note, that the interpolation is pretty good:

exact[x_, y_] := x^2 + y^2
int[1.1, 1.1] == exact[1.1, 1.1]
(* =&gt; True *)


Or better yet (thanks @rcollyer):

(int[1.1, 1.1] - exact[1.1, 1.1])/exact[1.1, 1.1]
(* 1.83508*10^-16 *)




Update Leonid's comment below pointed out that the accuracy of Interpolation will be worse with an unstructured grid. For example:

dataDelete = Delete[data, RandomInteger[{1, Length[data]}]]
intD = Interpolation[dataDelete]


Then,

(intD[1.1, 1.1] - exact[1.1, 1.1])/exact[1.1, 1.1]
(* ==&gt; 0.0743802 *)


which is worse. It seems particularly bad close to the origin:

Plot3D[(intD[x, y] - exact[x, y])/ exact[x, y], {x, -10, 10}, {y, -10, 10}]