1. Problem: Given sets $A = \{1, 3, 4, a\}$ and $B = \{2, 5, b, c\}$ where $a, b, c$ are digits, and the intersection $A \cap B$ has 2 elements. Also, $a + b + c = 22$. We need to find which of the sums 12, 13, or 14 can be the sum of elements in $B \setminus A$. 2. Understanding the problem: The intersection $A \cap B$ has 2 elements, meaning exactly two elements are common in both sets. Since $A$ and $B$ have digits and $a, b, c$ are unknown digits, the common elements must be among these. 3. Since $A = \{1, 3, 4, a\}$ and $B = \{2, 5, b, c\}$, the only way for $A \cap B$ to have 2 elements is if two of $a, b, c$ are equal to elements in the other set. 4. Elements in $A$ are 1, 3, 4, and $a$. Elements in $B$ are 2, 5, $b$, and $c$. The fixed elements 1, 3, 4 are not in $B$ and 2, 5 are not in $A$. So the intersection must come from $a$ matching either $b$ or $c$, and $b$ and $c$ matching $a$ or each other. 5. Since $A \cap B$ has 2 elements, and fixed elements do not overlap, the intersection must be $a$ equals one of $b$ or $c$, and the other of $b$ or $c$ equals an element in $A$ (which can only be $a$). So $a$ must be equal to one of $b$ or $c$, and the other must be equal to $a$ as well, which is impossible unless $b = c = a$. 6. But $b$ and $c$ are digits, so $a = b = c$ to have two common elements. But then $a + b + c = 3a = 22$ which is not an integer solution since $22/3$ is not an integer. 7. Alternatively, the intersection elements could be $a$ and one of the fixed elements 1, 3, or 4 if $b$ or $c$ equals that fixed element. But $B$ only has 2, 5, $b$, $c$, so $b$ or $c$ could be 1, 3, or 4. 8. Let's consider $b$ or $c$ equals one of 1, 3, or 4 to create intersection with $A$. 9. Suppose $b = 3$ and $a = 1$, then $A = \{1, 3, 4, 1\} = \{1, 3, 4\}$ and $B = \{2, 5, 3, c\}$. Intersection is $\{3, 1\}$ if $c = 1$ or $c = 4$. 10. Let's try to find all possible $a, b, c$ satisfying $a + b + c = 22$ and $|A \cap B| = 2$. 11. Since $A \cap B$ has 2 elements, and fixed elements 1, 3, 4 are in $A$, and 2, 5 in $B$, the intersection elements must be among $a, b, c$ and fixed elements. 12. Let's consider the intersection elements are $a$ and one of the fixed elements 1, 3, or 4, which must be in $B$ as $b$ or $c$. 13. So, $a$ is in $B$ as $b$ or $c$, and one of 1, 3, 4 is in $B$ as $b$ or $c$. 14. Since $B$ has $b$ and $c$, and fixed 2 and 5, the two intersection elements are $a$ and one of 1, 3, 4. 15. So $B = \{2, 5, a, d\}$ where $d$ is one of 1, 3, 4. 16. Then $A = \{1, 3, 4, a\}$ and $B = \{2, 5, a, d\}$ with $d \in \{1, 3, 4\}$ and $d \neq a$. 17. The intersection is $\{a, d\}$, so size 2. 18. Now, $a + b + c = 22$ and $b, c$ are $a$ and $d$ in some order. 19. So $a + b + c = a + a + d = 2a + d = 22$. 20. Since $a, d$ are digits (0-9), find integer solutions to $2a + d = 22$ with $d \in \{1, 3, 4\}$. 21. Try $d=1$: $2a + 1 = 22 \Rightarrow 2a = 21 \Rightarrow a = 10.5$ (not digit). 22. Try $d=3$: $2a + 3 = 22 \Rightarrow 2a = 19 \Rightarrow a = 9.5$ (not digit). 23. Try $d=4$: $2a + 4 = 22 \Rightarrow 2a = 18 \Rightarrow a = 9$ (valid digit). 24. So $a=9$, $d=4$, and $b, c$ are $9$ and $4$ in some order. 25. Then $B = \{2, 5, 9, 4\}$ and $A = \{1, 3, 4, 9\}$. 26. Now, $B \setminus A = \{2, 5\}$ because $9$ and $4$ are in $A$. 27. Sum of $B \setminus A = 2 + 5 = 7$, which is not among 12, 13, or 14. 28. Check if $b$ and $c$ can be other digits to get sums 12, 13, or 14. 29. Since $a=9$, $b=4$, $c=9$ or $b=9$, $c=4$, sum is 22. 30. $B \setminus A$ is $\{2, 5\}$ sum 7. 31. Try other possibilities: if $a=8$, $d=6$ (not in $\{1,3,4\}$), no. 32. So only $a=9$, $d=4$ works. 33. Now, consider $B \setminus A = \{2, 5, b, c\} \setminus \{1, 3, 4, a\}$. 34. Since $a=9$, $b=4$, $c=9$, $B = \{2, 5, 4, 9\}$ and $A = \{1, 3, 4, 9\}$. 35. $B \setminus A = \{2, 5\}$ sum 7. 36. None of 12, 13, 14. 37. Try if $a=7$, $d=8$ (not in $\{1,3,4\}$), no. 38. Try if $a=6$, $d=10$ no. 39. So only $a=9$, $d=4$. 40. Conclusion: None of 12, 13, or 14 can be the sum of $B \setminus A$. 41. But the problem asks which of 12, 13, 14 can be the sum of $B \setminus A$. 42. Let's check if $B \setminus A$ can be $\{b, c\}$ only, excluding 2 and 5. 43. If $b$ and $c$ are not in $A$, then $B \setminus A$ includes $b$ and $c$ plus 2 and 5 if they are not in $A$. 44. Since 2 and 5 are not in $A$, they are always in $B \setminus A$. 45. So sum of $B \setminus A$ is $2 + 5 +$ sum of $b$ and $c$ not in $A$. 46. Since $a=9$, $b=4$, $c=9$, $b$ and $c$ are in $A$, so $B \setminus A = \{2, 5\}$ sum 7. 47. If $b$ or $c$ is not in $A$, sum increases. 48. Try $a=8$, $b=7$, $c=7$ sum 22, intersection 2 elements? 49. $A=\{1,3,4,8\}$, $B=\{2,5,7,7\}$ intersection is empty, no. 50. Try $a=7$, $b=8$, $c=7$ sum 22, intersection? 51. $A=\{1,3,4,7\}$, $B=\{2,5,8,7\}$ intersection $\{7\}$ only 1 element. 52. Try $a=5$, $b=8$, $c=9$ sum 22, intersection? 53. $A=\{1,3,4,5\}$, $B=\{2,5,8,9\}$ intersection $\{5\}$ only 1 element. 54. Try $a=4$, $b=9$, $c=9$ sum 22, intersection? 55. $A=\{1,3,4,4\} = \{1,3,4\}$, $B=\{2,5,9,9\}$ intersection empty. 56. Try $a=3$, $b=9$, $c=10$ no digit 10. 57. Try $a=1$, $b=10$, $c=11$ no. 58. So only $a=9$, $d=4$ works. 59. Sum of $B \setminus A$ is 7. 60. None of 12, 13, 14. 61. Therefore, none of the given sums 12, 13, or 14 can be the sum of $B \setminus A$ under the given conditions. Final answer: None of I, II, or III can be the sum of $B \setminus A$.