Calculus (3rd Edition)

Since the normal vector to the plane $\frac{1}{2}x+2y-z=5$ is $\langle\frac{1}{2},2,-1 \rangle$ and the normal vector to the plane $3x+12y-6z=1$ is $\langle 3,12,-6 \rangle$, one can easily see that $$\langle 3,12,-6 \rangle=6\langle\frac{1}{2},2,-1 \rangle,$$ That is, the normal vectors are parallel and hence the planes are parallel.