Answer
a) $z=6-3 r \cos \theta-2r \sin \theta $
b) $z^2=1+r^2$
Work Step by Step
a) In the cylindrical co-ordinates system, we have $r^2=x^2+y^2 \implies r=\sqrt{x^2+y^2}$ and $x=r \cos \theta ; y=r \sin \theta; z=z$
Thus, we have
$3x+2y+z=6$
Thus, it can be written as:
$3 r \cos \theta+2r \sin \theta +z=6$ or, $z=6-3 r \cos \theta-2r \sin \theta $
b) We are given that $-x^2-y^2+z^2=1$
This can be written as:
$-r^2+z^2=1$
Hence, $z^2=1+r^2$
