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(A) For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{

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$k = 6$

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(A) For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.Given equations are $kx + 3y - (k - 3) = 0$ and $12x + ky - k = 0$.Here,$a_1 = k, b_1 = 3, c_1 = -(k - 3)$ and $a_2 = 12, b_2 = k, c_2 = -k$.Applying the condition: $\frac{k}{12} = \frac{3}{k} = \frac{-(k - 3)}{-k}$.From $\frac{k}{12} = \frac{3}{k}$,we get $k^2 = 36$,so $k = \pm 6$.From $\frac{3}{k} = \frac{k - 3}{k}$,we get $3k = k(k - 3) \implies 3k = k^2 - 3k \implies k^2 - 6k = 0 \implies k(k - 6) = 0$. Thus,$k = 0$ or $k = 6$.The common value satisfying both conditions is $k = 6$.

For what value of $k$ will the following pair of linear equations have infinitely many solutions?
$kx + 3y - (k - 3) = 0$
$12x + ky - k = 0$

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For what value of $k$ will the following pair of linear equations have infinitely many solutions?$kx + 3y - (k - 3) = 0$$12x + ky - k = 0$