For what value of k, the system of equations $3x - ky - 3 = 0$ and $2x - 3y - 4 = 0$ has no solution?

A system of two linear equations has no solution when the lines are parallel. This occurs when the lines have equal slopes but different y-intercepts.

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Converting the first equation $3x - ky - 3 = 0$ to slope-intercept form:

$ky = 3x - 3$

$y = \frac{3}{k}x - \frac{3}{k}$

The slope is $\frac{3}{k}$ and the y-intercept is $-\frac{3}{k}$.

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Converting the second equation $2x - 3y - 4 = 0$ to slope-intercept form:

$3y = 2x - 4$

$y = \frac{2}{3}x - \frac{4}{3}$

The slope is $\frac{2}{3}$ and the y-intercept is $-\frac{4}{3}$.

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For the lines to be parallel, the slopes must be equal:

$\frac{3}{k} = \frac{2}{3}$

$3 \times 3 = k \times 2$

$9 = 2k$

$k = \frac{9}{2}$

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When $k = \frac{9}{2}$, the y-intercepts are:

First equation: $-\frac{3}{9/2} = -\frac{3 \times 2}{9} = -\frac{2}{3}$

Second equation: $-\frac{4}{3}$

Since $-\frac{2}{3} \neq -\frac{4}{3}$, the lines are parallel with different y-intercepts, confirming no solution exists.

Therefore, $k = \frac{9}{2}$.