Answer
FALSE
Work Step by Step
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, nor are they orthogonal or perpendicular.
Hence, the statement is false.
