To determine the values of $k$ for which the given pair of linear equations will have infinitely many solutions, we need to check when the two equations are dependent, meaning they represent the same line.

The given equations are:

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

First, we rewrite both equations in the standard form $ax + by + c = 0$.

1. The first equation: $kx + 3y - k + 3 = 0 \quad \text{or} \quad kx + 3y + (3 - k) = 0$

   Comparing with the standard form, we have: $a_1 = k, \quad b_1 = 3, \quad c_1 = 3 - k$

2. The second equation: $12x + ky - k = 0$

   Here, we have: $a_2 = 12, \quad b_2 = k, \quad c_2 = -k$

For the system to have infinitely many solutions, the ratios of the corresponding coefficients must be equal, i.e.,

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Substituting the values:

$\frac{k}{12} = \frac{3}{k} = \frac{3-k}{-k}$

Let's solve these equations:

1. $\frac{k}{12} = \frac{3}{k}$

   Cross-multiplying gives: $k^2 = 36 \quad \Rightarrow \quad k = \pm 6$

2. $\frac{3}{k} = \frac{3-k}{-k}$

   Cross-multiplying gives: $-3k = k(3 - k) \quad \Rightarrow \quad -3k = 3k - k^2$ Simplifying: $k^2 - 6k = 0 \quad \Rightarrow \quad k(k - 6) = 0$ This gives: $k = 0 \quad \text{or} \quad k = 6$

Thus, the values of $k$ that satisfy both conditions are $k = 6$.

Hence, the pair of linear equations will have infinitely many solutions for $\boxed{k = 6}$.

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Would you like more details or have any questions? Here are some related questions to consider:

1. How do we determine if a pair of linear equations have no solution?
2. What happens if the ratio of coefficients is only partially satisfied?
3. How can you verify the solution geometrically?
4. What is the significance of the determinant in solving systems of equations?
5. Can we solve the equations graphically to find the value of $k$?

Tip: When working with systems of equations, always remember to check all conditions to ensure the correct number of solutions (one, none, or infinitely many).