A plane $\mathrm{E}_{1}$ makes intercepts $1,-3,4$ on the coordinate axes
Equation of plane is $\frac{x}{1}+\frac{y}{-3}+\frac{z}{4}=1 \Rightarrow 12 x-4 y+3 z=12$
d.r.s. are $12,-4,3$
Since required plane is parallel to given plane, normal vector $\bar{n}$ to required plane is
$\overline{\mathrm{n}}=12 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
The vector equation of the plane passing through $(2,6,-8)$ is
$\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}$, where $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$
$\bar{a} \cdot \bar{n}=(12)(2)-(4)(6)+(3)(-8)=24-24-24=-24$
$\therefore$ Required equation is $\overline{\mathrm{r}} \cdot(12 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=-24$
$\therefore$ Cartesion form of equation is $12 x-4 y+3 z+24=0$
$\therefore \frac{x}{1}-\frac{y}{3}+\frac{z}{4}+2=0$