Question

Find the equation of the plane which is parallel to the plane $x+5y-4z+5=0$ and the sum of where intercept on the co-ordinate axes is $15$.

Answer 1

Plane $:x+5y-4z+5=0$

A plane parallel to above plane will be of the form,

$x+5y-4z=k$

$x,y,z$ intercepts are $k,\frac{k}{5},\frac{-k}{4}$

$\therefore k+\frac{k}{5}-\frac{k}{4}=15\Rightarrow k=\frac{300}{9}$

$\therefore x+5y-4z=\frac{300}{9}$ is the equation of plane.