Four distinct points are marked in a line. For each point, the sum of the distances from said point to the other three is calculated; getting in total 4 numbers.

Decide whether these 4 numbers can be, in some order:
a) $29,29,35,37$
b) $28,29,35,37$
c) $28,34,34,37$
1. **Define the points and distances:**
   Let the four points be \( A, B, C, D \) in order from left to right. Define the distances between consecutive points as follows:
   \[
   AB = x, \quad BC = y, \quad CD = z
   \]

2. **Calculate the sum of distances from each point:**
   - For point \( A \):
     \[
     \text{Sum of distances} = AB + AC + AD = x + (x + y) + (x + y + z) = 3x + 2y + z
     \]
   - For point \( B \):
     \[
     \text{Sum of distances} = BA + BC + BD = x + y + (y + z) = x + 2y + z
     \]
   - For point \( C \):
     \[
     \text{Sum of distances} = CA + CB + CD = (x + y) + y + z = x + 2y + z
     \]
   - For point \( D \):
     \[
     \text{Sum of distances} = DA + DB + DC = (x + y + z) + (y + z) + z = x + 2y + 3z
     \]

3. **Identify the necessary conditions:**
   - From the calculations, we see that the sums of distances from \( B \) and \( C \) are equal:
     \[
     \text{Sum from } B = \text{Sum from } C = x + 2y + z
     \]
   - The sums of distances from \( A \) and \( D \) are:
     \[
     \text{Sum from } A = 3x + 2y + z
     \]
     \[
     \text{Sum from } D = x + 2y + 3z
     \]

4. **Check each option:**
   - **Option (a) \( 29, 29, 35, 37 \):**
     - We need two equal numbers, which are \( 29 \) and \( 29 \). Assign these to \( B \) and \( C \):
       \[
       x + 2y + z = 29
       \]
     - The remaining sums are \( 35 \) and \( 37 \). Assign these to \( A \) and \( D \):
       \[
       3x + 2y + z = 35
       \]
       \[
       x + 2y + 3z = 37
       \]
     - Solve the system of equations:
       \[
       x + 2y + z = 29 \quad \text{(1)}
       \]
       \[
       3x + 2y + z = 35 \quad \text{(2)}
       \]
       \[
       x + 2y + 3z = 37 \quad \text{(3)}
       \]
     - Subtract (1) from (2):
       \[
       3x + 2y + z - (x + 2y + z) = 35 - 29
       \]
       \[
       2x = 6 \implies x = 3
       \]
     - Subtract (1) from (3):
       \[
       x + 2y + 3z - (x + 2y + z) = 37 - 29
       \]
       \[
       2z = 8 \implies z = 4
       \]
     - Substitute \( x = 3 \) and \( z = 4 \) back into (1):
       \[
       3 + 2y + 4 = 29
       \]
       \[
       2y = 22 \implies y = 11
       \]
     - Therefore, \( x = 3 \), \( y = 11 \), \( z = 4 \) satisfies all conditions. Thus, option (a) is possible.

   - **Option (b) \( 28, 29, 35, 37 \):**
     - We need two equal numbers, but there are no two equal numbers in this set. Thus, option (b) is not possible.

   - **Option (c) \( 28, 34, 34, 37 \):**
     - We need two equal numbers, which are \( 34 \) and \( 34 \). Assign these to \( B \) and \( C \):
       \[
       x + 2y + z = 34
       \]
     - The remaining sums are \( 28 \) and \( 37 \). Assign these to \( A \) and \( D \):
       \[
       3x + 2y + z = 28
       \]
       \[
       x + 2y + 3z = 37
       \]
     - Solve the system of equations:
       \[
       x + 2y + z = 34 \quad \text{(1)}
       \]
       \[
       3x + 2y + z = 28 \quad \text{(2)}
       \]
       \[
       x + 2y + 3z = 37 \quad \text{(3)}
       \]
     - Subtract (1) from (2):
       \[
       3x + 2y + z - (x + 2y + z) = 28 - 34
       \]
       \[
       2x = -6 \implies x = -3
       \]
     - Subtract (1) from (3):
       \[
       x + 2y + 3z - (x + 2y + z) = 37 - 34
       \]
       \[
       2z = 3 \implies z = 1.5
       \]
     - Substitute \( x = -3 \) and \( z = 1.5 \) back into (1):
       \[
       -3 + 2y + 1.5 = 34
       \]
       \[
       2y = 35.5 \implies y = 17.75
       \]
     - Since \( x = -3 \), \( y = 17.75 \), \( z = 1.5 \) are not valid distances (distances cannot be negative), option (c) is not possible.

Conclusion:
Only option (a) \( 29, 29, 35, 37 \) is possible.

The final answer is \( \boxed{ 29, 29, 35, 37 } \)