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.
![A+B>C\\A+C>B\\B+C>A](https://tex.z-dn.net/?f=A%2BB%3EC%5C%5CA%2BC%3EB%5C%5CB%2BC%3EA)
<h2>First Option: {6, 10, 12}</h2>
Let A = 6, B = 10, and C = 12:
![6+10>12\\16>12\ \textsf{true}\\\\6+12>10\\18>10\ \textsf{true}\\\\12+10>6\\22>6\ \textsf{true}](https://tex.z-dn.net/?f=6%2B10%3E12%5C%5C16%3E12%5C%20%5Ctextsf%7Btrue%7D%5C%5C%5C%5C6%2B12%3E10%5C%5C18%3E10%5C%20%5Ctextsf%7Btrue%7D%5C%5C%5C%5C12%2B10%3E6%5C%5C22%3E6%5C%20%5Ctextsf%7Btrue%7D)
<h2>Second Option: {5, 7, 10}</h2>
Let A = 5, B = 7, and C = 10
![5+7>10\\12>10\ \textsf{true}\\\\5+10>7\\15>7\ \textsf{true}\\\\10+7>5\\17>5\ \textsf{true}](https://tex.z-dn.net/?f=5%2B7%3E10%5C%5C12%3E10%5C%20%5Ctextsf%7Btrue%7D%5C%5C%5C%5C5%2B10%3E7%5C%5C15%3E7%5C%20%5Ctextsf%7Btrue%7D%5C%5C%5C%5C10%2B7%3E5%5C%5C17%3E5%5C%20%5Ctextsf%7Btrue%7D)
<h2>Third Option: {4, 4, 9}</h2>
Let A = 4, B = 4, and C = 9
![4+4>9\\8>9\ \textsf{false}\\\\4+9>4\\13>4\ \textsf{true}\\\\4+9>4\\13>4\ \textsf{true}](https://tex.z-dn.net/?f=4%2B4%3E9%5C%5C8%3E9%5C%20%5Ctextsf%7Bfalse%7D%5C%5C%5C%5C4%2B9%3E4%5C%5C13%3E4%5C%20%5Ctextsf%7Btrue%7D%5C%5C%5C%5C4%2B9%3E4%5C%5C13%3E4%5C%20%5Ctextsf%7Btrue%7D)
<h2>Fourth Option: {2, 3, 3}</h2>
Let A = 2, B = 3, and C = 3
![2+3>3\\5>3\ \textsf{true}\\\\2+3>3\\5>3\ \textsf{true}\\\\3+3>2\\6>2\ \textsf{true}](https://tex.z-dn.net/?f=2%2B3%3E3%5C%5C5%3E3%5C%20%5Ctextsf%7Btrue%7D%5C%5C%5C%5C2%2B3%3E3%5C%5C5%3E3%5C%20%5Ctextsf%7Btrue%7D%5C%5C%5C%5C3%2B3%3E2%5C%5C6%3E2%5C%20%5Ctextsf%7Btrue%7D)
<h2>Conclusion:</h2>
Since the third option had an inequality that was false, <u>the third option cannot be a triangle.</u>