$l: x=0,y=t,z=t$ and $\pi:6x+2y-2z=3$ find if they are parallel and how is above the other.
So I took the dot product $(0,1,1)\cdot(6,2,-2)=0$ so they are parallel. To test how is above/below I have set $x=0$ and looked at the $z$ component.
for the line I got $z=-y$ and for the plane I got $z=y-\frac{3}{2}$, which has different values for positive and negative $y$ values.
Is there a different approach?
On
Based on my comments ... The plane containing the line $x=0,y=t,z=t$ has an equation of $6x+2y-2z =0$, which means $z_2={3x+y}$ (the value of $z$ for plane 2). The original plane had an equation of $6x+2y-2z=3$, or $z_1={3x+y} -{3\over2}$. Since $z_2> z_1$, the plane containing the line is above the original plane.
Take any point $\boldsymbol{p}=(x_p,y_p,z_p)$ on the line and find it's signed distance to the plane
$$ d = \frac{6 x_p + 2 y_p +(-2) z_p - 3}{\sqrt{(6)^2+(2)^2+(-2)^2}} $$
if it is positive then the point is in the same side as the plane normal direction. If it is negative it is in the opposite side.
In your case, the line passes through the origin $(0,0,0)$ yielding a negative $d$.