For which value of $k$ does the pair $kx+3y=k-3$ and $12x+ky=k$ have no solution?

No solution needs a₁/a₂ = $b$₁/$b$₂ ≠ c₁/c₂. From $k/12=3/k$ we get $k^2=36$, so $k$ = ±6. If $k=6$ all three ratios equal 1/2 (infinitely many solutions). If $k=-6$ the first two ratios are −1/2 while $(k-3)/k=-9/-6=3/2$, so $k=-6$ gives no solution.