Answer
$$3x+2y+z=3, \quad 3x+2y+z=2.$$
(Other answers are possible.)
Work Step by Step
The equation of any plane is given by
$$ a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0.$$
Since $ n=\langle 3,2,1\rangle $, then we can arbitrarily choose the planes as follows
$$3(x-1)+2(y-0)+(z-0)=0, \quad 3(x-0)+2(y-1)+(z-0)=0.$$
Note that we could have chosen any other vector, as long as does not pass through $(0,0,0)$.
By simplification, we have
$$3x+2y+z=3, \quad 3x+2y+z=2.$$
