Question

Find the equation of the plane through the point p=(5,5,4)p=(5,5,4) and parallel to the plane 5x+2y−z=−6.

Answer 1

If the required plane is parallel to the plane
Π : 5x+2y-z=-6
then they both have the normal vector <5,2,-1>

Given the required plane passes through the point (5,5,4), the new plane is defined by
5(x-5)+2(y-5)-1(z-4)=0
=>
plane: 5x+2y-z=31