I'm asked to find an equation of the line in $R^3$ that contains the point $P_0(0,1,2)$, is parallel to the plane $x+y+z=2$ $(1)$ and perpendicular to the line $x=1+t, y=1-t,z=2t$ $(2)$
By substituting the point on the parametric equations we now have
$l:$ $x=ta_1$, $y=1+ta_2$, $z=2+ta_3$.
From $(1)$ we have: $<\mathbf n=(1,1,1), \mathbf a=(a_1,a_2,a_3)>=0 \implies a_1+a_2+a_2=0$
Similarly from $(2) : a_1-a_2+2a_3=0$
It seems to me that we should have all the information we need by now but there seems to be something missing as we cannot find the equation yet. How de we proceed?
You were not far from the result.
A parametric description of this line is
$$\pmatrix{x\\y\\z}=P_0+tV_1=\pmatrix{0\\1\\2}+t\pmatrix{3\\-1\\-2}$$
Why this expression of $V_1$ ?
Because vector $V_1$ ,which has to be orthogonal to the normal vector to the plane $\pmatrix{1\\1\\1}$ and to the direction vector $\pmatrix{1\\-1\\2}$ has thus to be proportional to the cross product
$$\tag{*}\pmatrix{1\\1\\1} \times \pmatrix{1\\-1\\2}=\pmatrix{3\\-1\\-2}.$$
Edit: Why where you close to the result ? Your relationships
$$\begin{cases}(2) : a_1+a_2+a_3=0\\(2) : a_1-a_2+2a_3=0\end{cases}$$
constitute an underdetermined system of 2 equations with 3 unknowns ; for this kind of system, there is a standard procedure.
Write this system as a square system of 2 equations with $2$ unknowns, byconsidering the third variable $a_3$ as a parameter, i.e. write it under the form
$$\begin{cases}(1) &:& a_1+a_2=-a_3\\(2)& :& a_1-a_2=-2a_3\end{cases}$$
The solutions to this parametric system are:
$$\begin{cases}a_1=\frac12(-3a_3)\\a_2=\frac12(a_3)\end{cases}$$
Thus the general solution is
$$\pmatrix{a_1\\a_2\\a_3}=\pmatrix{\frac12(-3a_3)\\\frac12(a_3)\\a_3}$$
which is in fact proportional to the cross product we have found in (*)