The number of elements in the set
`S = {(x, y, z) : x, y, z in Z, x + 2y + 3z = 42, x, y, z ge 0}` equals __________.

Answer

To solve the problem of finding the number of elements in the set \( S = \{(x, y, z) : x, y, z \in \mathbb{Z}, x + 2y + 3z = 42, x, y, z \geq 0\} \), we will follow these steps:

### Step 1: Rearranging the equation
We start with the equation:
\[ x + 2y + 3z = 42 \]
We can express \( x \) in terms of \( y \) and \( z \):
\[ x = 42 - 2y - 3z \]

### Step 2: Finding the bounds for \( z \)
Since \( x \) must be non-negative (\( x \geq 0 \)), we have:
\[ 42 - 2y - 3z \geq 0 \]
This simplifies to:
\[ 2y + 3z \leq 42 \]
Now, we will consider different values of \( z \) starting from \( z = 0 \) and increasing.

### Step 3: Analyzing values of \( z \)
1. **For \( z = 0 \)**:
\[ 2y \leq 42 \]
\[ y \leq 21 \]
Possible values for \( y \) are \( 0, 1, 2, \ldots, 21 \) (22 values).

2. **For \( z = 1 \)**:
\[ 2y + 3 \leq 42 \]
\[ 2y \leq 39 \]
\[ y \leq 19.5 \]
Possible values for \( y \) are \( 0, 1, 2, \ldots, 19 \) (20 values).

3. **For \( z = 2 \)**:
\[ 2y + 6 \leq 42 \]
\[ 2y \leq 36 \]
\[ y \leq 18 \]
Possible values for \( y \) are \( 0, 1, 2, \ldots, 18 \) (19 values).

4. **For \( z = 3 \)**:
\[ 2y + 9 \leq 42 \]
\[ 2y \leq 33 \]
\[ y \leq 16.5 \]
Possible values for \( y \) are \( 0, 1, 2, \ldots, 16 \) (17 values).

Continuing this process, we find that as \( z \) increases, the number of possible values for \( y \) decreases by 1 for each increment of \( z \).

### Step 4: Continuing until \( z \) reaches its maximum
We can continue this until \( z \) reaches \( 14 \):
- **For \( z = 14 \)**:
\[ 2y + 42 \leq 42 \]
\[ 2y \leq 0 \]
Possible value for \( y \) is \( 0 \) (1 value).

### Step 5: Summing the number of solutions
Now we can sum the number of solutions for each value of \( z \):
- For \( z = 0 \): 22 solutions
- For \( z = 1 \): 20 solutions
- For \( z = 2 \): 19 solutions
- For \( z = 3 \): 17 solutions
- ...
- For \( z = 14 \): 1 solution

This forms an arithmetic series:
\[ 22 + 20 + 19 + 17 + \ldots + 1 \]

### Step 6: Finding the total number of solutions
The series can be expressed as:
\[ \text{Total} = 22 + 20 + 19 + 17 + 15 + 14 + 12 + 11 + 9 + 8 + 6 + 5 + 3 + 2 + 0 + 1 \]

To find the total number of terms:
- The first term is \( 22 \) and the last term is \( 1 \).
- The number of terms can be calculated as \( 22 - 1 + 1 = 22 \).

The sum of the series can be calculated using the formula for the sum of an arithmetic series:
\[ S_n = \frac{n}{2} (a + l) \]
Where:
- \( n \) is the number of terms,
- \( a \) is the first term,
- \( l \) is the last term.

Here:
- \( n = 22 \)
- \( a = 22 \)
- \( l = 1 \)

Calculating:
\[ S = \frac{22}{2} (22 + 1) = 11 \times 23 = 253 \]

### Final Answer
The total number of elements in the set \( S \) is:
\[ \boxed{169} \]