For what value of k, the following pair of linear equations have infinitely many solutions : kx + 3y – (k – 3) = 0 12x + ky – k = 0

The correct answer is (a).

For a system of linear equations to have infinitely many solutions, the two equations must be equal. In this case, the equations are:

$$kx + 3y – (k – 3) = 0$$
$$12x + ky – k = 0$$

Subtracting the second equation from the first equation, we get:

$$-9x + 2y = 3$$

Dividing both sides by 2, we get:

$$-4.5x + y = \frac{3}{2}$$

We can see that this equation is always true, regardless of the value of $k$. Therefore, the system of equations has infinitely many solutions for any value of $k$.

The other options are incorrect because they do not make the system of equations equal. For example, if $k = 3$, the first equation becomes $3x + 9y – 6 = 0$ and the second equation becomes $12x + 3y – 3 = 0$. These equations are not equal, so the system of equations does not have infinitely many solutions.