Orthogonal Projection of $v$ on sub-space $U$ 
What is the orthogonal projection of $v = (1,5,-10)$ on the sub-space $U = Sp\{(5,-2,1),(1,2,-1)\}$?

Well, I managed to compute it on the way I've been taught by the book, but it doesn't seem to work. Any chance you guys can solve it? If needed, I can provide the way I tried to solve it.
 A: Hint: 
1) Compute an orthonormal base $u_1,u_2$ of $U$ using Gramm-SChmidt
2) Consider the projector $p_U(x) = \sum_{i=1}^2 <u_i,x>u_i$
3) Compute $p_U(v)$
So let's do it:
Note that $$\langle \begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix}5\\ -2 \\ 1 \end{pmatrix}\rangle = 0 \quad \text{ and }\quad \|\begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}\|_2 = \sqrt{6}, \ \|\begin{pmatrix} 5\\ -2 \\ 1 \end{pmatrix}\|_2 = \sqrt{30}.$$
So we may choose 
$$ u_1 = \frac{1}{\sqrt{30}}\begin{pmatrix} 5\\ -2 \\ 1 \end{pmatrix}\quad \text{ and }\quad u_2=\frac{1}{\sqrt{6}}\begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}$$
and thus $p_U(v)$, the projection of $v$ on $U$ is given by
$$p_U(v) = \langle u_1,v\rangle u_1 +\langle u_2,v\rangle u_2= \frac{-15}{\sqrt{30}}u_1+\frac{1}{\sqrt{6}}u_2 = 
\begin{pmatrix}\frac{-75}{30}+\frac{1}{6} \\ \frac{30}{30}+\frac{2}{6}  \\ \frac{-15}{30}-\frac{1}{6}\end{pmatrix}=\frac{1}{3}\begin{pmatrix}-7 \\ 4\\ -2\end{pmatrix}.$$
A: Another approach is to compute the normal to the subspace using the cross product:
$$
\begin{pmatrix}5,-2,1\end{pmatrix}\times\begin{pmatrix}1,2,-1\end{pmatrix}=\begin{pmatrix}0,6,12\end{pmatrix}
$$
Then subtract the component of $\begin{pmatrix}1,5,-10\end{pmatrix}$ in that direction:
$$
\begin{align}
&\begin{pmatrix}1,5,-10\end{pmatrix}-\overbrace{\frac{\begin{pmatrix}1,5,-10\end{pmatrix}\cdot\begin{pmatrix}0,6,12\end{pmatrix}}{\begin{pmatrix}0,6,12\end{pmatrix}\cdot\begin{pmatrix}0,6,12\end{pmatrix}}\begin{pmatrix}0,6,12\end{pmatrix}}^{\begin{array}{c}\text{orthogonal component of $\begin{pmatrix}1,5,-10\end{pmatrix}$}\\\text{ in the direction of $\begin{pmatrix}0,6,12\end{pmatrix}$}\end{array}}\\
&=\begin{pmatrix}1,5,-10\end{pmatrix}-\frac{-90}{180}\begin{pmatrix}0,6,12\end{pmatrix}\\
&=\begin{pmatrix}1,5,-10\end{pmatrix}+\begin{pmatrix}0,3,6\end{pmatrix}\\[3pt]
&=\begin{pmatrix}1,8,-4\end{pmatrix}
\end{align}
$$

Check
$$
\begin{pmatrix}1,8,-4\end{pmatrix}=\frac72\begin{pmatrix}1,2,-1\end{pmatrix}-\frac12\begin{pmatrix}5,-2,1\end{pmatrix}
$$
so the computed vector is in the given span.
