Hello, Way Below Zero!
The plane x + 2y + z = 1 has the normal vector <1,2,1>.
We want the line through A(3,1,2) perpendicular to the plane.
We have point A(3,1,2) and direction <1,2,1>,
so the parametric equations are:
x = 3 + t, y = 1 + 2t + z = 2 + t
Find the intersection of this line and the plane:
(3 + t) + 2(1 + 2t) + (2 + t) = 1
Hence, t = -1.
They intersect at B(2,-1,1).
Here's where I used some intuition instead of formulas.
Going from A(3,1,2) to B(2,-1,1), we note the changes in the coordinates:
x decreases by 1, y decreases by 2, z decreases by 1.
If we continue in the same manner to point C, we get C(1,-3,0).
Therefore, the image of A is (1,-3,0).