Find the value of k for which the system of equations kx + 3y = k - 3 and 12x + ky = k has no solution.

For no solution, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Here, $\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$. Checking the constant ratio: for $k = 6$, $\frac{c_1}{c_2} = \frac{6-3}{6} = \frac{3}{6} = \frac{1}{2}$, while $\frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2}$, which leads to infinitely many solutions. For $k = -6$, $\frac{c_1}{c_2} = \frac{-6-3}{-6} = \frac{-9}{-6} = \frac{3}{2}$, whereas $\frac{a_1}{a_2} = \frac{-6}{12} = -\frac{1}{2}$ (not equal), giving no solution. Thus, $k = -6$.