C

Step-by-step Explanation:

Let $A$ and $B$ be the magnitudes of two non-zero vectors $\vec{A}$ and $\vec{B}$, respectively. Since they are non-zero, $A > 0$ and $B > 0$.

The magnitude of their resultant vector $\vec{C}$ when the angle between them is $\theta = 120^\circ$ is given by:
$$C = \sqrt{A^2 + B^2 + 2AB \cos(120^\circ)}$$

Since $\cos(120^\circ) = -\frac{1}{2}$, substituting this value gives:
$$C = \sqrt{A^2 + B^2 - AB}$$

Squaring both sides:
$$C^2 = A^2 + B^2 - AB$$

Now, consider the square of $|A - B|$:
$$|A - B|^2 = A^2 + B^2 - 2AB$$

Comparing $C^2$ and $|A - B|^2$ by taking their difference:
$$C^2 - |A - B|^2 = (A^2 + B^2 - AB) - (A^2 + B^2 - 2AB) = AB$$

Since $A > 0$ and $B > 0$, the product $AB > 0$.
Therefore:
$$C^2 > |A - B|^2 \implies C > |A - B|$$

Thus, $C$ must be greater than $|A - B|$.

Correct Option: C