Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Preliminary Questions - Page 684: 1

$3x+4y-z=0.$

Since the plane is parallel to $3x+4y-z=5$, then its normal vector will be $\langle 3,4,-1\rangle $, and since the plane passes through the origin, the equation of the plane is given by $$3(x-0)+4(y-0)-(z-0)=0\Longrightarrow 3x+4y-z=0.$$