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(B) For a system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have no solution,the condition is $\frac{a_1}{a_2} = \frac

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$-\frac{4}{3}$

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(B) For a system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$.Given equations are $2x - ky + 3 = 0$ and $3x + 2y - 1 = 0$.Here,$a_1 = 2, b_1 = -k, c_1 = 3$ and $a_2 = 3, b_2 = 2, c_2 = -1$.Applying the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2}$:$\frac{2}{3} = \frac{-k}{2}$Multiplying both sides by $2$,we get $4 = -3k$.Therefore,$k = -\frac{4}{3}$.Checking the condition $\frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$:$\frac{-(-4/3)}{2} = \frac{4/3}{2} = \frac{2}{3}$.Since $\frac{c_1}{c_2} = \frac{3}{-1} = -3$,and $\frac{2}{3} 
eq -3$,the condition holds true for $k = -\frac{4}{3}$.

Find the value of $k$ for the following pair of linear equations such that the system has no solution:
$2x - ky + 3 = 0$
$3x + 2y - 1 = 0$

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Find the value of $k$ for the following pair of linear equations such that the system has no solution:$2x - ky + 3 = 0$$3x + 2y - 1 = 0$