## Calculus: Early Transcendentals 8th Edition

Given: $6x-2y+4z=1$ The normal vector of the plane is $6x-2y+4z=1$ $\lt 6,-2, 4\gt$ Divide by 2 $\lt 3,-1, 2\gt$ Thus the normal vectors are not parallel; either they are orthogonal or perpendicular. Hence, the statement is false.