Suppose $S = \{1, 2, 3, x\}$ is a set with four distinct real numbers for which the difference between the largest and smallest values of $S$ is equal to the sum of elements of $S.$ What is the value of $x?$
$$
\mathrm a. ~ {-1}\qquad \mathrm b.~{-3/2}\qquad \mathrm c. ~{-2} \qquad \mathrm d. ~{-2/3} \qquad \mathrm e. ~{-3}
$$
1. Given the set \( S = \{1, 2, 3, x\} \) with four distinct real numbers, we need to find the value of \( x \) such that the difference between the largest and smallest values of \( S \) is equal to the sum of the elements of \( S \).

2. The sum of the elements of \( S \) is:
   \[
   1 + 2 + 3 + x = 6 + x
   \]

3. We need to consider different cases for the value of \( x \) to determine the largest and smallest values in \( S \).

4. **Case 1: \( x \geq 3 \)**
   - The largest value in \( S \) is \( x \) and the smallest value is \( 1 \).
   - The difference between the largest and smallest values is:
     \[
     x - 1
     \]
   - According to the problem, this difference should equal the sum of the elements:
     \[
     x - 1 = 6 + x
     \]
   - Solving for \( x \):
     \[
     x - 1 = 6 + x \implies -1 = 6 \quad \text{(impossible)}
     \]
   - Therefore, \( x \geq 3 \) is not a valid case.

5. **Case 2: \( 3 > x \geq 1 \)**
   - The largest value in \( S \) is \( 3 \) and the smallest value is \( 1 \).
   - The difference between the largest and smallest values is:
     \[
     3 - 1 = 2
     \]
   - According to the problem, this difference should equal the sum of the elements:
     \[
     2 = 6 + x
     \]
   - Solving for \( x \):
     \[
     2 = 6 + x \implies x = 2 - 6 \implies x = -4
     \]
   - Since \( -4 \) is not in the interval \( 3 > x \geq 1 \), this case is also not valid.

6. **Case 3: \( x < 1 \)**
   - The largest value in \( S \) is \( 3 \) and the smallest value is \( x \).
   - The difference between the largest and smallest values is:
     \[
     3 - x
     \]
   - According to the problem, this difference should equal the sum of the elements:
     \[
     3 - x = 6 + x
     \]
   - Solving for \( x \):
     \[
     3 - x = 6 + x \implies 3 - 6 = 2x \implies -3 = 2x \implies x = -\frac{3}{2}
     \]
   - Since \( -\frac{3}{2} < 1 \), this case is valid.

Therefore, the value of \( x \) that satisfies the given conditions is \( x = -\frac{3}{2} \).

The final answer is \( \boxed{-\frac{3}{2}} \)