The required plane is parallel to 3x + 2y - z=7, so required plane and the given plane must have the same normal vector.

Vector normal to the plane 3x + 2y - z=7 is

We know that, equation of plane perpendicular to a given direction & passing through a given point is given by,

Here, it is given that, the plane passes through (2, - 1, 1) so in this case, in vector form, can be denoted as,

Equation of the required plane is,

3(x - 2) + 2(y + 1) - (z - 1)=0

3x - 6 + 2y + 2 - z + 1=0

3x + 2y - z=3

Hence, the equation of the plane passing through (2, –1, 1) and parallel to the plane 3x + 2y–z=7 is 3x + 2y - z=3.