Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 684: 25

$$ x+z=3.$$

Since the normal vector $ n $ is $\langle 1,0,1\rangle $ and, using the fact that the plane passes through $(-2,-3,5)$, the equation of the plane is given by $$(x+2)+0(y+3)+(z-5)=0\Longrightarrow x+z-3=0$$ Hence the equation is given by $$ x+z=3.$$