The given equation is \(3x+2y+z=6\)

we know \(x=r\cos\theta,y=r\sin\theta\) and z=z

\(\therefore\) we have \(3r\cos\theta+2r\sin\theta+z=6\)

Result:

\(3r\cos\theta+2r\sin\theta+z=6\)

we know \(x=r\cos\theta,y=r\sin\theta\) and z=z

\(\therefore\) we have \(3r\cos\theta+2r\sin\theta+z=6\)

Result:

\(3r\cos\theta+2r\sin\theta+z=6\)