Answer
$x+3y+4z=34$
Work Step by Step
The normal to the plane is: $n=\lt 1,3,4 \gt$
We know that the standard equation of a plane passing through the point $(x_0,y_0,z_0)$ is written as: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Then for point $(2,4,5)$, we have
$(1)(x-2)+(3)(y-4)+(4)(z-5)=0$
or, $x-2+3y-12+4z-20=0$
or, $x+3y+4z=34$
