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 find 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 of the plane:
\[ x + 2y + 3z = 15 \]
We can rearrange this equation to express \( x + 2y \) in terms of \( z \):
\[ x + 2y = 15 - 3z \]

### Step 2: Finding Positive Integral Solutions
We will consider different integer values for \( z \) and find corresponding positive integral values for \( x \) and \( y \).

#### Case 1: \( z = 1 \)
Substituting \( z = 1 \):
\[ x + 2y = 15 - 3(1) = 12 \]
Now we need to find positive integral solutions for \( x + 2y = 12 \):
- If \( x = 2 \), then \( 2y = 10 \) → \( y = 5 \)
- If \( x = 4 \), then \( 2y = 8 \) → \( y = 4 \)
- If \( x = 6 \), then \( 2y = 6 \) → \( y = 3 \)
- If \( x = 8 \), then \( 2y = 4 \) → \( y = 2 \)
- If \( x = 10 \), then \( 2y = 2 \) → \( y = 1 \)

Total solutions for \( z = 1 \): **5 solutions**.

#### Case 2: \( z = 2 \)
Substituting \( z = 2 \):
\[ x + 2y = 15 - 3(2) = 9 \]
Finding positive integral solutions for \( x + 2y = 9 \):
- If \( x = 1 \), then \( 2y = 8 \) → \( y = 4 \)
- If \( x = 3 \), then \( 2y = 6 \) → \( y = 3 \)
- If \( x = 5 \), then \( 2y = 4 \) → \( y = 2 \)
- If \( x = 7 \), then \( 2y = 2 \) → \( y = 1 \)

Total solutions for \( z = 2 \): **4 solutions**.

#### Case 3: \( z = 3 \)
Substituting \( z = 3 \):
\[ x + 2y = 15 - 3(3) = 6 \]
Finding positive integral solutions for \( x + 2y = 6 \):
- If \( x = 2 \), then \( 2y = 4 \) → \( y = 2 \)
- If \( x = 4 \), then \( 2y = 2 \) → \( y = 1 \)

Total solutions for \( z = 3 \): **2 solutions**.

#### Case 4: \( z = 4 \)
Substituting \( z = 4 \):
\[ x + 2y = 15 - 3(4) = 3 \]
Finding positive integral solutions for \( x + 2y = 3 \):
- If \( x = 1 \), then \( 2y = 2 \) → \( y = 1 \)

Total solutions for \( z = 4 \): **1 solution**.

#### Case 5: \( z = 5 \)
Substituting \( z = 5 \):
\[ x + 2y = 15 - 3(5) = 0 \]
This case does not yield any positive integral solutions since both \( x \) and \( y \) must be positive.

### Step 3: Summing the Solutions
Now, we sum the total number of solutions from all cases:
- For \( z = 1 \): 5 solutions
- For \( z = 2 \): 4 solutions
- For \( z = 3 \): 2 solutions
- For \( z = 4 \): 1 solution
- For \( z = 5 \): 0 solutions

Total \( n = 5 + 4 + 2 + 1 + 0 = 12 \).

### Step 4: Finding \( n/2 \)
Finally, we find \( n/2 \):
\[ \frac{n}{2} = \frac{12}{2} = 6 \]

### Final Answer
Thus, the value of \( n/2 \) is \( \boxed{6} \).