Using the condition that if two lines $l_1 x+m_1 y+n_1=0$ and $l_2 x+m_2 y+n_2=0$ are conjugate w.r.t. parabola $y^2=4 a x$, then
$$
l_1 n_2+l_2 n_1=2 a m_1 m_2
$$
Given conjugate lines are $2 x+3 y+12=0$ and $x-y+4 \lambda=0$ and equation of parabola is $y^2=8 x$.
Here, $l_1=2, m_1=3, n_1=12 ; l_2=1, m_2=-1$, $n_2=4 \lambda$ and $a=2$
$\therefore$ From Eq. (i),
$$
\begin{gathered}
2 \times 4 \lambda+1 \times 12=2 \times 2 \times 3 \times(-1) \\
8 \lambda=-12-12 \Rightarrow \lambda=-3
\end{gathered}
$$