The value of p for the following pair of linear equations ( p - 3)x + 3y = p , px + py =12 have infinitely many solutions is ………

Answer

To find the value of \( p \) for which the pair of linear equations

\[ (p - 3)x + 3y = p \]
\[ px + py = 12 \]

has infinitely many solutions, we can follow these steps:

### Step 1: Write the equations in standard form
We rewrite both equations in the standard form \( Ax + By + C = 0 \).

1. For the first equation:
\[ (p - 3)x + 3y - p = 0 \]
This gives us:
- \( A_1 = p - 3 \)
- \( B_1 = 3 \)
- \( C_1 = -p \)

2. For the second equation:
\[ px + py - 12 = 0 \]
This gives us:
- \( A_2 = p \)
- \( B_2 = p \)
- \( C_2 = -12 \)

### Step 2: Set up the condition for infinitely many solutions
For the two equations to have infinitely many solutions, the following condition must hold:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

### Step 3: Set up the equations based on the condition
From the above condition, we can set up the following equations:

1. From \( \frac{A_1}{A_2} = \frac{B_1}{B_2} \):
\[ \frac{p - 3}{p} = \frac{3}{p} \]

2. From \( \frac{A_1}{A_2} = \frac{C_1}{C_2} \):
\[ \frac{p - 3}{p} = \frac{-p}{-12} \]

### Step 4: Solve the first equation
From the first equation:
\[ \frac{p - 3}{p} = \frac{3}{p} \]
Cross-multiplying gives:
\[ (p - 3) \cdot p = 3 \cdot p \]
This simplifies to:
\[ p^2 - 3p = 3p \]
\[ p^2 - 6p = 0 \]
Factoring out \( p \):
\[ p(p - 6) = 0 \]
Thus, \( p = 0 \) or \( p = 6 \).

### Step 5: Check the second equation
Now, we check the second equation:
\[ \frac{p - 3}{p} = \frac{-p}{-12} \]
Cross-multiplying gives:
\[ (p - 3)(-12) = p^2 \]
This simplifies to:
\[ -12p + 36 = p^2 \]
Rearranging gives:
\[ p^2 + 12p - 36 = 0 \]
However, we already found \( p = 6 \) from the first equation.

### Conclusion
The value of \( p \) for which the pair of linear equations has infinitely many solutions is:
\[ \boxed{6} \]