This is very peculiar. You say that you know that any plane can be written as Ax+ By+ Cz+ D= 0. It has already been pointed out that you an divide through by any number, say, D, to reduce to three unknown constants, A'x+ B'y+ C'z+ 1= 0 (with A'
= A/D etc.). If the line given by the vector formula r= (3,2,1)+t(1,0,2) or parametric equations x= 3+ t, y= 2, z= 1+ 2t. In particular, taking t= 0, (3, 2, 1) is a point in the plane so we must have 3A'+ 2B'+ C'+ 1= 0.
Also, taking t= 1, (3+ 1, 2+ 0, 1+ 2)= (4, 2, 3) is also a point in the plane. so we must have 4A'+ 2B'+ 3C'+ 1= 0.
The last condition is that the plane is parallel to the z-axis. That means that its normal vector, <A, B, C> or <A', B', C'> is normal to <0, 0, 1>. Their dot product gives C= C'= 0 and so we have the two equations -3A'- 2B'- 1= 0 and 4A'+ 2B'+ 1= 0.
By the way, I find it simpler to use the fact that if a plane has normal vector <A, B, C> and contains point [itex](x_0, y_0, z_0)[/itex], then the plane has equation [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex].
