Given, conjugate lines are
$2 x+3 y+12=0$ ...(i)
and $\quad x-y+k=0$ ...(ii)
We know, that two lines are said to be conjugate with respect to a curve, if each passes through the pole of the polar of that curve. Let $\left(x_1, y_1\right)$ be the pole of parabola
$y^2=8 x$
It's polar is $\quad y y_1=4\left(x+x_1\right)$
$\Rightarrow \quad 4 x-y_1 y+4 x_1=0$
$\Rightarrow \quad 2 x-\left(\frac{y_1}{2}\right) y+2 x_1=0$ ...(iii)
On comparing Eqs. (i) and (iii),
we get $\frac{-y_1}{2}=3 \Rightarrow y_1=-6$ and $2 x_1=12 \Rightarrow x_1=6$
$\therefore \quad$ Pole $\left(x_1, y_1\right)=(6,-6)$
Eq. (ii) also passes through pole $(6,-6)$
$\therefore \quad 6-(-6)+k=0$
$\Rightarrow \quad k=-12$