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(A) For a pair of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \

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$6$

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(A) For a pair of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ 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 $kx + 3y = (k - 3)$ and $12x + ky = k$.Here,$a_1 = k, b_1 = 3, c_1 = k - 3$ and $a_2 = 12, b_2 = k, c_2 = k$.Applying the condition: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$.Taking the first two parts: $\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$.Taking the last two parts: $\frac{3}{k} = \frac{k - 3}{k} \implies 3k = k(k - 3) \implies 3k = k^2 - 3k \implies k^2 - 6k = 0 \implies k(k - 6) = 0$.This gives $k = 0$ or $k = 6$.For the condition to hold for all ratios,we must satisfy both equations. The common value is $k = 6$.

Find the value of $k$ for which the following pair of linear equations has infinitely many solutions:
$kx + 3y = k - 3$
$12x + ky = k$

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Find the value of $k$ for which the following pair of linear equations has infinitely many solutions:$kx + 3y = k - 3$$12x + ky = k$