Linear algebra question about a plane in $\mathbb{R}^3$ not through the origin

Let $\mathcal{P}$ be the plane in $\mathbb{R}^3$ containing the points $A=(0,1,1)$, $B=(1,-1,1)$, and $C=(1,0,1)$. Let $\mathcal{L}$ be the line perpendicular to $\mathcal{P}$, containing the point $A$. Let $U=(1,-1,0)$. Find the following.

1. a scalar equation for the plane $\mathcal{P}$.
2. a vector equation for the line $\mathcal{L}$.
3. The point on the line $\mathcal{L}$ closest to the point $U$.

My approach was to translate the plane in such a way that it ends up crossing the origin, and then translating all the points referred to the same way. Then I translated the answers back.

Is this the correct method? If so, could somebody please give a detailed answer, as there is no solution in the book for this problem. If not so, why is this wrong?

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(1) A scalar equation looks like $ax+by+cz=d$, so you need to find $a,b,c$ such that $A, B$, and $C$ solve the equation. Hint: it turns out that $d$ can just be $1$, assuming that the plane containing $A,B$, and $C$ is not already going through the origin. So this is just three equations in three variables.
Your approach isn't wrong, but you really don't need the translation, given that you can just solve a linear system to get $(a,b,c)$.
(2) Once you have that plane, $(a,b,c)$ is perpendicular to it. So you just have to send a line through $A$ going that direction.
(3) This is just a matter of projecting $U$ onto $\mathcal{L}$, which is really just projecting onto $(a,b,c)$. But $\mathcal{L}$ doesn't go through the origin, so you have to make some point on $\mathcal{L}$ the origin, like $A$.