Two points on the line $( L$ say $) \frac{ x }{3}=\frac{ y }{2}, z =1$ are
(0,0,1)$\&(3,2,1)$
So dr's of the line is $<3,2,0>$
Line passing through $(1,2,1),$ parallel to $L$ and
coplanar with given plane is
$\vec{ r }=\hat{ i }+2 \hat{ j }+\hat{ k }+ t (3 \hat{ i }+2 j ), t \in R (-2,0,1)$ satisfies
the line $($ for $t=-1)$
$\Rightarrow (-2,0,1)$ lies on given plane.
Answer of the question is (2)
We can check other options by finding equation of plane
Equation plane :
$\begin{vmatrix}x-1&y-2&z-1\\ 1+2&2-0&1-1\\ 2+2&1-0&2-1\end{vmatrix} = 0$
$\Rightarrow 2(x-1)-3(y-2)-5(z-1)=0$
$\Rightarrow 2 x-3 y-5 z+9=0$