A plane \(\mathrm{E}_{1}\) makes intercepts \(1,-3,4\) on the co-ordinate axes. The equation of a plane
parallel to plane \(\mathrm{E}_{1}\) and passing through \((2,6,-8)\) is

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 \(12 x-4 y+3 z+24=0\)
\(\therefore \frac{x}{1}-\frac{y}{3}+\frac{z}{4}+2=0\)