Given a line g: $\bar x = P + \lambda \bar t$ and the plane $E$: $\bar x = A + \alpha \bar a + \beta \bar b $ . Determine the location of the point $P'\neq P$, which lies on g and has the same distance to the plane $E$ as the point $P$
So I figured that vectors $\bar a$ and $\bar b$ should belong to $E$, hence their cross product will be normal to the plane, which I have found $\bar n=(0,-2,-2)$. Using $\bar n$ and the coordinates of point A I came up with the equation of the plane in different form $E=\bar n(\bar x - A)=0 \rightarrow 2x_2 -2x_3-6=0$ And then I got lost, the formula for distance between plane and point would give $3$ unknowns in one equation. Clearly, I should look into some other direction.