Find the line that goes through $\ (0,1,2) $, is parallel to the plane $\ x+y+z-2=0 $, and is perpendicular to $\ r(t) = (1+t,1-t,2t) $.
I understand that the line is perpendicular to the vectors $\ (1,-1,2)$ and $(1,1,1) $, but I'd like some guidance as to what I should do.
To find a vector in $\mathbb R^3$ perpendicular to two other linearly independent vectors, take the cross product, which can be computed as a formal determinant:
$$\left|\begin{matrix} \hat i &\hat j & \hat k\\ 1 & -1 & 2 \\ 1 & 1 & 1 \end{matrix}\right| = -3\hat i + \hat j + 2\hat k .$$
If you like, you could verify using the dot product that $(-3,1,2)$ is perpendicular to the other two vectors.
So the line you're seeking is given by $l(t)=(0,1,2)+t(-3,1,2)=(-3t,1+t,2+2t)$.