Two water pipelines are represented by the equations $kx + 3y +1 = 0$ and $2x + y + 3 = 0$. For what value of $k$, the pipelines cross each other?

$k≠6$

The correct answer is Option (3) → $k≠6$

For two straight lines to cross each other, they must not be parallel, i.e., their slopes must be different.

Given equations:

1. $kx + 3y + 1 = 0$

$3y = -kx - 1 \Rightarrow y = -\frac{k}{3}x - \frac{1}{3}$

Slope $m_1 = -\frac{k}{3}$​

2. $2x + y + 3 = 0$

$y = -2x – 3$

Slope $m_2 = -2$

For the lines to cross:

$m_1 \ne m_2$

$-\frac{k}{3} \ne -2$

$k \ne 6$