The value of k for which the pair of linear equations `kx+3y+(3-k)=0 and 12x+ky-k=0` have infinitely many solution is :

Options

A. `-6`
B. 12
C. `-12`
D. `+6`.

Answer

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

1. \( kx + 3y + (3 - k) = 0 \)
2. \( 12x + ky - k = 0 \)

have infinitely many solutions, we need to use the condition that the ratios of the coefficients of \( x \), \( y \), and the constant terms must be equal.

### Step 1: Identify coefficients
From the first equation, we can identify the coefficients:
- \( a_1 = k \)
- \( b_1 = 3 \)
- \( c_1 = 3 - k \)

From the second equation, we can identify the coefficients:
- \( a_2 = 12 \)
- \( b_2 = k \)
- \( c_2 = -k \)

### Step 2: Set up the ratio condition
For the equations to have infinitely many solutions, the following must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

This gives us three equations to solve:
1. \( \frac{k}{12} = \frac{3}{k} \)
2. \( \frac{3}{k} = \frac{3 - k}{-k} \)

### Step 3: Solve the first equation
Starting with the first equation:
\[ \frac{k}{12} = \frac{3}{k} \]
Cross-multiplying gives:
\[ k^2 = 36 \]
Taking the square root of both sides, we find:
\[ k = 6 \quad \text{or} \quad k = -6 \]

### Step 4: Solve the second equation
Now, we solve the second equation:
\[ \frac{3}{k} = \frac{3 - k}{-k} \]
Cross-multiplying gives:
\[ 3(-k) = (3 - k)k \]
Simplifying, we get:
\[ -3k = 3k - k^2 \]
Rearranging terms leads to:
\[ k^2 - 6k = 0 \]
Factoring out \( k \):
\[ k(k - 6) = 0 \]
Thus, \( k = 0 \) or \( k = 6 \).

### Step 5: Identify common solutions
From the first equation, we found \( k = 6 \) or \( k = -6 \). From the second equation, we found \( k = 0 \) or \( k = 6 \). The common solution is:
\[ k = 6 \]

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