Official solution
The equation of plane
(x + y + z - 1) + \[\lambda \] (2x + 3y - z + 4) = 0
\[ \Rightarrow \] (1 + 2\[\lambda \])x + (1 + 3\[\lambda \])y + (1 - \[\lambda \])z + 4\[\lambda \] - 1 = 0
As plane is parallel to y axis so the normal vector of plane and dot product of \[\widehat j\] is zero.
\[ \therefore \] 1 + 3\[\lambda \] = 0
\[ \Rightarrow \] \[\lambda \] = - \[{1 \over 3}\]
\[ \therefore \] So the equation of the plane is
x(1 - \[{2 \over 3}\]) + (1 - \[{3 \over 3}\]) y + (1 + \[{1 \over 3}\]) - \[{4 \over 3}\] - 1 = 0
\[ \Rightarrow \] x (\[{1 \over 3}\]) + z(\[{4 \over 3}\]) - \[{7 \over 3}\] = 0
\[ \Rightarrow \] x + 4z - 7 = 0
By checking each options you can see only point (3, 2, 1) lies on the plane.