How to compute the projection of a vector on a plane Can someone check whether my work is correct or not?

Compute the projection of $(1,1,1)$ onto the plane that passes through the points $(1,0,-1), (3,7,-3), (-2,-1,2)$. 

My attempt: Let $u = (1,1,1)$. I first try to find a vector normal to the plane given by the three points. By  computing a cross product, I got $n = (19,0,19) = (1,0,1)$.
The projection of $u$ onto the plane is  $$\frac{u.n}{||n||^2}u.$$ Is that correct? 
 A: A plane is of the form $a_1 x_1 + a_2 x_2 + a_3 x_3 = b$. Since we are given three points on the plane, we obtain a system of $3$ linear equations in $a_1, a_2, a_3, b$. One solution would be $a_1 = a_3 = 1$, $a_2 = 0$ and $b = 0$. Hence, an equation of the plane that passes through the three given points is 
$$x_1 + x_3 = 0$$
though any multiple of this equation would also do. Note that this plane also passes through the origin. The vector that is normal to this plane is 
$$\mathrm{n} = \begin{bmatrix}1\\ 0 \\ 1\end{bmatrix}$$
We can decompose the point to be projected, $\mathrm{u}$, as the sum of a vector normal to the plane and a vector parallel to the plane
$$\mathrm{u} = \begin{bmatrix}1\\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}1\\ 0 \\ 1\end{bmatrix} + \begin{bmatrix} 0\\ 1 \\ 0\end{bmatrix} = \mathrm{n} + \begin{bmatrix} 0\\ 1 \\ 0\end{bmatrix}$$
Projecting $\mathrm{u}$ onto the plane we obtain only
$$\begin{bmatrix} 0\\ 1 \\ 0\end{bmatrix}$$
as the projection of a vector normal to the plane onto the same plane is the zero vector.
