The value of 'k' such that the straight line $2x+3y+4+k(6x-y+12)=0$  is parallel to $X-$axis is

$\begin{array}{l}2x+3y+4+k(6x-y+12)=0\\ \Rightarrow 2x+3y+4+6kx-yk+12k=0\\ \Rightarrow (2+6k)x+(3-k)y+4+12k=0\end{array}$

Slope of parallel to $X-$axis  is '0'

$\begin{array}{l}\Rightarrow \frac{-(2+6k)}{3-k}=0\\ \Rightarrow -2-6k=0\Rightarrow -2-6k=0\\ \Rightarrow 6k=-2\\ \therefore k=\frac{-1}{3}\end{array}$