Step 1: Start from the no-solution condition in cross-multiplied form.
For $kx + 3y = k - 3$ and $12x + ky = k$, a necessary condition for the lines to be parallel is $a_1 b_2 = a_2 b_1$, that is \[ k \times k = 12 \times 3 \]
Step 2: Solve this equation for k.
\[ k^2 = 36 \quad \Rightarrow \quad k = 6 \text{ or } k = -6 \]
Step 3: Rule out the value that makes the lines coincide instead of merely parallel.
For genuinely no solution, the lines must be parallel but distinct, so $\frac{c_1}{c_2}$ must differ from $\frac{a_1}{a_2}$. Checking $k = 6$: the ratios $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{3}{6} = \frac{1}{2}$ all come out equal, so this actually gives infinitely many solutions, not none. That rules out $k = 6$.
Step 4: Confirm k = -6 works.
For $k = -6$: $\frac{a_1}{a_2} = -\frac{1}{2}$, $\frac{b_1}{b_2} = -\frac{1}{2}$, but $\frac{c_1}{c_2} = \frac{-9}{-6} = \frac{3}{2}$, which is different, exactly the pattern needed for no solution.
\[ \boxed{k = -6} \]