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I have a discretized 3D surface for which I want to compute piece-wise linear hat functions. I assumed these functions are of the following form:

$$phi = ax + by + cz + d$$

with the property of $phi_i(x_j) = delta_{ij}$.

For a general 3D surface, the number of neighbors for each vertex is different, ranging from 3 to 6 or even more. For example, in the figure below, 3D vertex $i$ has 6 neighbors. This means:

$$
phi_{17}(x_{17}) = ax_{17} + by_{17} + cz_{17} + d_{17} = 1,\
phi_{17}(x_{11}) = ax_{11} + by_{11} + cz_{11} + d_{11} = 0,\
phi_{17}(x_{12}) = ax_{12} + by_{12} + cz_{12} + d_{12} = 0,\
phi_{17}(x_{13}) = ax_{13} + by_{13} + cz_{13} + d_{13} = 0,\
phi_{17}(x_{14}) = ax_{14} + by_{14} + cz_{14} + d_{14} = 0,\
phi_{17}(x_{15}) = ax_{15} + by_{15} + cz_{15} + d_{15} = 0,\
phi_{17}(x_{16}) = ax_{16} + by_{16} + cz_{16} + d_{16} = 0.\
$$

The system of equations for vertex 17 is overdetermined (7 equations and 4 unknowns) and not necessarily lead to a set of coefficients, $(a, b, c, d)$, that satisfy $phi_i(x_j) = delta_{ij}$.

I was wondering how piece-wise linear hat functions should be computed for such surfaces to guarantee the hat function is $1$ at vertex $i$ and $0$ at all other vertices.

First define one triangle and interpolations as follows:
$$
begin{bmatrix} x \ y \ z end{bmatrix} =
begin{bmatrix} x_1 \ y_1 \ z_1 end{bmatrix} +
begin{bmatrix} x_2-x_1 \ y_2-y_1 \ z_2-z_1 end{bmatrix} xi +
begin{bmatrix} x_3-x_1 \ y_3-y_1 \ z_3-z_1end{bmatrix} eta
qquad mbox{with:} quad begin{cases} xi > 0 \ eta > 0 \ xi+eta < 1 end{cases}
$$
And quite in general:
$$
phi = phi_1 + (phi_2-phi_1),xi + (phi_3-phi_1),eta
$$
From this reference we infer that:
$$
begin{cases}
xi = [ (y_3 - y_1).(x - x_1) - (x_3 - x_1).(y - y_1) ] / Delta \
eta = [ (x_2 - x_1).(y - y_1) - (y_2 - y_1).(x - x_1) ] / Delta
end{cases}
$$
Where $Delta$ is twice the area of a triangle.

The Finite Element shape functions thus are, still for one triangle:
$$
N_1 = 1-xi-eta quad ; quad N_2 = xi quad ; quad N_3 = eta
$$ $$
x = N_1x_1+N_2x_2+N_3x_3 \
y = N_1y_1+N_2y_2+N_3y_3 \
z = N_1z_1+N_2z_2+N_3z_3 \
phi = N_1phi_1+N_2phi_2+N_3phi_3
$$
Now specify for the 6 triangles in you mesh and you're done, for $,xi = eta = 0,$ eventually:

1. $(1)rightarrow(17) ;,; (2)rightarrow(11) ;,; (3)rightarrow(12)$


2. $(1)rightarrow(17) ;,; (2)rightarrow(12) ;,; (3)rightarrow(13)$


3. $(1)rightarrow(17) ;,; (2)rightarrow(13) ;,; (3)rightarrow(14)$


4. $(1)rightarrow(17) ;,; (2)rightarrow(14) ;,; (3)rightarrow(15)$


5. $(1)rightarrow(17) ;,; (2)rightarrow(15) ;,; (3)rightarrow(16)$


6. $(1)rightarrow(17) ;,; (2)rightarrow(16) ;,; (3)rightarrow(11)$