You are trying to maximize a function of the form $f(x, y, z) = ax + by + cz$, where $a$, $b$, and $c$ are constants. You know that $f(3, 1, 1) > f(2, 1, 1)$, $f(2, 2, 3) > f(2, 3, 4)$, and $f(3, 3, 4) > f(3, 3, 3)$. For $-5 \le x,y,z \le 5$, what value of $(x,y,z)$ maximizes the value of $f(x, y, z)$? Give your answer as an ordered triple.
1. We start by analyzing the given inequalities to determine the signs of the constants \(a\), \(b\), and \(c\).

2. The inequality \(f(3, 1, 1) > f(2, 1, 1)\) can be written as:
   \[
   a \cdot 3 + b \cdot 1 + c \cdot 1 > a \cdot 2 + b \cdot 1 + c \cdot 1
   \]
   Simplifying this, we get:
   \[
   3a + b + c > 2a + b + c
   \]
   \[
   a > 0
   \]
   This tells us that \(a\) is positive.

3. The inequality \(f(3, 3, 4) > f(3, 3, 3)\) can be written as:
   \[
   a \cdot 3 + b \cdot 3 + c \cdot 4 > a \cdot 3 + b \cdot 3 + c \cdot 3
   \]
   Simplifying this, we get:
   \[
   3a + 3b + 4c > 3a + 3b + 3c
   \]
   \[
   c > 0
   \]
   This tells us that \(c\) is positive.

4. The inequality \(f(2, 2, 3) > f(2, 3, 4)\) can be written as:
   \[
   a \cdot 2 + b \cdot 2 + c \cdot 3 > a \cdot 2 + b \cdot 3 + c \cdot 4
   \]
   Simplifying this, we get:
   \[
   2a + 2b + 3c > 2a + 3b + 4c
   \]
   \[
   2b + 3c > 3b + 4c
   \]
   \[
   2b + 3c - 3b - 4c > 0
   \]
   \[
   -b - c > 0
   \]
   \[
   -b > c
   \]
   Since we already know \(c > 0\), this implies:
   \[
   b < 0
   \]
   This tells us that \(b\) is negative.

5. To maximize the value of \(f(x, y, z) = ax + by + cz\), we need to maximize the positive contributions and minimize the negative contributions. Since \(a\) and \(c\) are positive, we should choose the maximum values for \(x\) and \(z\), which are 5. Since \(b\) is negative, we should choose the minimum value for \(y\), which is -5.

6. Therefore, the values of \(x\), \(y\), and \(z\) that maximize \(f(x, y, z)\) are:
   \[
   (x, y, z) = (5, -5, 5)
   \]

The final answer is \(\boxed{(5, -5, 5)}\).