Number of points having positive intergal co - ordinate lying on the plane `x+2y+3z = 15` is n, then `n/2` is equal to

Answer

To solve the problem of finding the number of points with positive integral coordinates lying on the plane given by the equation \(x + 2y + 3z = 15\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ x + 2y + 3z = 15 \]
We can rearrange it to express \(x\) in terms of \(y\) and \(z\):
\[ x = 15 - 2y - 3z \]

### Step 2: Finding Constraints for \(x\)
Since \(x\) must be a positive integer, we need:
\[ 15 - 2y - 3z > 0 \]
This simplifies to:
\[ 2y + 3z < 15 \]

### Step 3: Finding Possible Values of \(z\)
Next, we will consider different values of \(z\) and find the corresponding values of \(y\).

1. **For \(z = 1\)**:
\[ 2y + 3(1) < 15 \implies 2y < 12 \implies y < 6 \]
Possible values for \(y\) are \(1, 2, 3, 4, 5\). This gives us 5 values for \(y\).

2. **For \(z = 2\)**:
\[ 2y + 3(2) < 15 \implies 2y < 9 \implies y < 4.5 \]
Possible values for \(y\) are \(1, 2, 3, 4\). This gives us 4 values for \(y\).

3. **For \(z = 3\)**:
\[ 2y + 3(3) < 15 \implies 2y < 6 \implies y < 3 \]
Possible values for \(y\) are \(1, 2\). This gives us 2 values for \(y\).

4. **For \(z = 4\)**:
\[ 2y + 3(4) < 15 \implies 2y < 3 \implies y < 1.5 \]
The only possible value for \(y\) is \(1\). This gives us 1 value for \(y\).

5. **For \(z = 5\)**:
\[ 2y + 3(5) < 15 \implies 2y < 0 \]
This is not possible since \(y\) must be positive.

### Step 4: Counting Total Solutions
Now, we sum the number of valid \((y, z)\) pairs:
- For \(z = 1\): 5 values of \(y\)
- For \(z = 2\): 4 values of \(y\)
- For \(z = 3\): 2 values of \(y\)
- For \(z = 4\): 1 value of \(y\)

Total number of points \(n\):
\[ n = 5 + 4 + 2 + 1 = 12 \]

### Step 5: Finding \(n/2\)
Finally, we calculate:
\[ \frac{n}{2} = \frac{12}{2} = 6 \]

### Conclusion
Thus, the value of \(n/2\) is:
\[ \boxed{6} \]