Question

Find the distance from the point q=(4,1,−5) to the plane 4x+3y+2z=9

Answer 1

The plane is given by
f(x,y,z)=4x+3y+2z=9
Since f(4,1,-5)=4*4+3*1+2(-5)=16+3-10=9,
point q lies on the plane, therefore the distance is zero.