I'm having trouble with this problem -
Let $Q = (-2, 3, 4)$, and let $P$ be the foot of the perpendicular from $Q$ to the plane through points $A = (0,1,1), B = (1,1,0)$ and $C = (1,0,3)$.
Then $\overrightarrow{QP}$ is the projection of $\overrightarrow{QA}$ onto a vector $$\mathbf{u} = \begin{pmatrix}u_1\\ u_2 \\ u_3 \end{pmatrix},$$such that $u_1 +u_2 + u_3 = -5.$ What is $\mathbf{u}$? I've tried so many things. So far, I've found the equation for the plane - $x+3y+z-4=0$. I also know that QP is projecting onto vector AP, but I can't find it.
Thank you so much!
HINT
The normal vector to the plane is
$$\vec n=BA\times CA=\begin{vmatrix}\hat i&\hat j&\hat k\\1&0&-1\\1&-1&2\end{vmatrix}=(-1,-3,-1)$$