So for this, we will be applying the Triangle Inequality Theorem, which states that the sum of 2 sides must be greater than the third side for it to be a triangle. If any inequality turns out to be false, then the set cannot be a triangle.

<h2>First Option: {6, 10, 12}</h2>
Let A = 6, B = 10, and C = 12:

<h2>Second Option: {5, 7, 10}</h2>
Let A = 5, B = 7, and C = 10

<h2>Third Option: {4, 4, 9}</h2>
Let A = 4, B = 4, and C = 9

<h2>Fourth Option: {2, 3, 3}</h2>
Let A = 2, B = 3, and C = 3

<h2>Conclusion:</h2>
Since the third option had an inequality that was false, <u>the third option cannot be a triangle.</u>