The foot of the perpendicular from the origin to the lines $x+y-4=0,x+5y-26=0,15x-27y-424=0$ lie on a line. The equation of the line is

$\begin{array}{l}(0,0)\&x+y-4=0\\ \therefore \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-(a{x}_1+b{y}_1+c)}{a^2+b^2}\\ \frac{h}{1}=\frac{k}{1}=\frac{-(-4)}{2}\\ \therefore (h,k)=(2,2)\\ (0,0)\&x+5y-26=0\\ \frac{h-0}{1}=\frac{k-0}{5}=\frac{-(-26)}{26}\\ \therefore (h,k)=(1,5)\end{array}$

The line equation of $(2,2),(1,5)$ is$y-5=\frac{3}{-1}(x-1)$

$\begin{array}{l}-y+5=3x-3\\ \therefore 3x+y-8=0\end{array}$