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(D) The given pair of linear equations is:$kx + 3y = (k - 3)$$12x + ky = k$Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$

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$-6$

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(D) The given pair of linear equations is:$kx + 3y = (k - 3)$$12x + ky = k$Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = k, b_1 = 3, c_1 = -(k - 3)$$a_2 = 12, b_2 = k, c_2 = -k$For the pair of linear equations to have no solution,the condition is:$\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$$\frac{k}{12} = \frac{3}{k} 
eq \frac{-(k - 3)}{-k}$Taking the first two parts:$\frac{k}{12} = \frac{3}{k} \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$Now,checking the condition $\frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$:If $k = 6$,then $\frac{3}{6} = \frac{1}{2}$ and $\frac{k-3}{k} = \frac{6-3}{6} = \frac{3}{6} = \frac{1}{2}$. Since $\frac{1}{2} = \frac{1}{2}$,$k = 6$ gives infinitely many solutions.If $k = -6$,then $\frac{3}{-6} = -\frac{1}{2}$ and $\frac{k-3}{k} = \frac{-6-3}{-6} = \frac{-9}{-6} = \frac{3}{2}$. Since $-\frac{1}{2} 
eq \frac{3}{2}$,the condition holds.Thus,the value of $k$ for which there is no solution is $-6$.

For what value$(s)$ of $k$ does the pair of equations $kx + 3y = k - 3$ and $12x + ky = k$ have no solution?

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