When defining functions, you mostly want to use SetDelayed (:=) and not Set (=). The difference between these two isn't something often see in other languages, but SetDelayed (:=) behaves more like you expect. When you use Set (=) the right hand side is evaluated first.

This is probably an abuse of error checking, but we can use Check to just return 0 whenever there's an error message.

safeExpression[x_?NumericQ, y_?NumericQ, z_?NumericQ] := Check[Abs[vHP] , 0]

Remember to include the arguments to the function when calling it. I've prepended "Quiet@" to stop the messages from appearing.

intensityTable = Table[If[ x^2 + y^2 + z^2 <= rMax2, Quiet@safeExpression[x, y, z], 0], {x, -a, a, 2 a/(n - 1)}, {y, -a, a, 2 a/(n - 1)}, {z, -a, a, 2 a/(n - 1)}] ;

This is a full programming language. So you can create conditional statements to handle errors like division by zero. I'm kinda guessing without seeing your code, but lets assume you're still working with this expression:

expression = 
 E^(I (-(x/2) - y/2 + z/2 - 3 ArcTan[(-(x/Sqrt[2]) + y/Sqrt[2])/(-(x/Sqrt[6]) - y/Sqrt[6] - Sqrt[2/3] z)]));

We can wrap a function around it that takes only numeric values. If it produces an Indeterminate value, 0 is returned instead.

safeExpression[x_?NumericQ, y_?NumericQ, z_?NumericQ] = (expression /. {Indeterminate -> 0})

So the function doesn't exist at a value. I assume them you want to use the limit at that point then? Does the limit exist?

To illustrate, let's look at the last expression. The problematic part is

ArcTan[(Sqrt[3] (x - y))/(x + y + 2 z)]

What value does this approach as x,y,and z approach 0? It depends entirely on how you approach 0. The limit doesn't exist.

Using the two argument form of ArcTan is often helpful, but usually where the limit exists. Replacing f/g with f/(g+.0001) is really just the same as choosing a random direction to take the limit from.

The point {0,0,0} isn't an isolated problem. The limit also doesn't exist if we let x and y approach some value "a" and let z approach "-a".

Limit[Limit[Limit[ArcTan[Sqrt[3] (x - y),  x + y + 2 z], {y -> a}], {z -> -a}], {x -> a}]
Pi/6

Limit[Limit[Limit[ArcTan[Sqrt[3] (x - y), x + y + 2 z], {x -> a}], {y -> a}], {z -> -a}]
Pi/2

So if you must evaluate this function at these points, replacing the functions value with the limit isn't going to do the trick. You have to decide what it means to evaluate the function at these points where it is not defined.