Find the closest point to $\mathbf{y}$ in the subspace $W$ spanned by $\mathbf{u}_{1}$ and $\mathbf{u}_{2}$
\[
\mathbf{y}=\left[\begin{array}{r}
17 \\
1 \\
5
\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{r}
1 \\
0 \\
-1
\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{l}
2 \\
1 \\
2
\end{array}\right]
\]
A. $\left[\begin{array}{r}-16 \\ -5 \\ -4\end{array}\right]$
B. $\left[\begin{array}{r}16 \\ 5 \\ 4\end{array}\right]$
C. $\left[\begin{array}{r}20 \\ 9 \\ 16\end{array}\right]$
D. $\left[\begin{array}{c}17 \\ 6 \\ 7\end{array}\right]$

Step 1 ：Calculate the dot products and the norms: \(\mathbf{y} \cdot \mathbf{u}_{1} = 17*1 + 1*0 + 5*(-1) = 12\), \(\mathbf{y} \cdot \mathbf{u}_{2} = 17*2 + 1*1 + 5*2 = 39\), \(\left\| \mathbf{u}_{1} \right\|^2 = 1^2 + 0^2 + (-1)^2 = 2\), \(\left\| \mathbf{u}_{2} \right\|^2 = 2^2 + 1^2 + 2^2 = 9\)

Step 2 ：Substitute these values into the formula: \(\operatorname{proj}_{W} \mathbf{y} = \frac{12}{2} \mathbf{u}_{1} + \frac{39}{9} \mathbf{u}_{2} = 6\mathbf{u}_{1} + \frac{13}{3}\mathbf{u}_{2}\)

Step 3 ：Calculate the projection: \(\operatorname{proj}_{W} \mathbf{y} = 6\left[\begin{array}{r}1 \\ 0 \\ -1\end{array}\right] + \frac{13}{3}\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right] = \left[\begin{array}{r}16 \\ \frac{13}{3} \\ 4\end{array}\right]\)

Step 4 ：So, the closest point to \(\mathbf{y}\) in the subspace \(W\) is \(\boxed{\text{B. }\left[\begin{array}{r}16 \\ 5 \\ 4\end{array}\right]}\)