Let $f: \mathbb{Z} \to \mathbb{Z}$ be a function satisfying the property that for any integers $a$, $b$, and $c$ such that $a + b + c = 0$, the equation

$$f(a) + f(b) + f(c) = a^2 + b^2 + c^2$$

holds. We need to determine all such functions $f$.

### Step 1: Understand the Condition

Since $a + b + c = 0$, we have $c = -(a+b)$. Substituting $c = -(a+b)$ into the given equation:

$$f(a) + f(b) + f(-(a+b)) = a^2 + b^2 + (-(a+b))^2.$$

Simplifying the right-hand side:

$$= a^2 + b^2 + (a+b)^2 = a^2 + b^2 + a^2 + 2ab + b^2 = 2a^2 + 2b^2 + 2ab.$$

Thus, the equation becomes:

$$f(a) + f(b) + f(-(a+b)) = 2a^2 + 2b^2 + 2ab.$$

### Step 2: Exploratory Substitution

Consider $a = x$, $b = 0$, and $c = -x$. Thus, we have:

$$f(x) + f(0) + f(-x) = x^2 + 0^2 + (-x)^2 = 2x^2.$$

### Step 3: Deriving the Form of $f$

By the above, we have:

$$f(x) + f(-x) = 2x^2 - f(0).$$

Now, observe the function:

We substitute $f(x) = x^2 + cx$ into the condition to verify it satisfies the equation:

$$f(x) + f(y) + f(-(x+y)) = (x^2 + cx) + (y^2 + cy) + (-(x+y))^2 - c(x+y).$$

Simplifying:

$$= x^2 + cx + y^2 + cy + (x+y)^2 + c(x+y).$$
$$= x^2 + y^2 + x^2 + 2xy + y^2 + cx + cy - cx - cy.$$
$$= 2x^2 + 2y^2 + 2xy.$$

Thus, the form $f(x) = x^2 + cx$ satisfies the condition.

### Conclusion

The solution to the functional equation is a function of the form:

$$f(x) = x^2 + cx \quad \text{for any integer } c.$$

Thus, all functions $f: \mathbb{Z} \to \mathbb{Z}$ satisfying the given property are:

$$\boxed{f(x) = x^2 + cx \text{ for any integer } c}.$$