Find the orthogonal projection of $\vec{v}=\left[\begin{array}{c}-1 \\ 11 \\ -5\end{array}\right]$ onto the subspace $V$ of $\mathbb{R}^{3}$ spanned by $\vec{x}=\left[\begin{array}{c}3 \\ 2 \\ -6\end{array}\right]$ and $\vec{y}=\left[\begin{array}{c}6 \\ -3 \\ 2\end{array}\right]$. (Note that the two vectors $\vec{x}$ and $\vec{y}$ are orthogonal to each other.) \[ \operatorname{proj}_{V}(\vec{v})=\left[\begin{array}{l} \square \\ \square \\ \square \end{array}\right] \]

Step 1 ：Given vectors are \(\vec{v} = \left[\begin{array}{c}-1 \\ 11 \\ -5\end{array}\right]\), \(\vec{x} = \left[\begin{array}{c}3 \\ 2 \\ -6\end{array}\right]\) and \(\vec{y} = \left[\begin{array}{c}6 \\ -3 \\ 2\end{array}\right]\)

Step 2 ：Calculate the dot products: \(\vec{v} \cdot \vec{x} = (-1*3) + (11*2) + (-5*-6) = 49\), \(\vec{v} \cdot \vec{y} = (-1*6) + (11*-3) + (-5*2) = -49\), \(\vec{x} \cdot \vec{x} = (3*3) + (2*2) + (-6*-6) = 49\), \(\vec{y} \cdot \vec{y} = (6*6) + (-3*-3) + (2*2) = 49\)

Step 3 ：Substitute these into the formula for orthogonal projection: \(\operatorname{proj}_{V}(\vec{v}) = \frac{49}{49} \vec{x} + \frac{-49}{49} \vec{y} = \vec{x} - \vec{y}\)

Step 4 ：Calculate \(\vec{x} - \vec{y} = \left[\begin{array}{c}3 \\ 2 \\ -6\end{array}\right] - \left[\begin{array}{c}6 \\ -3 \\ 2\end{array}\right] = \left[\begin{array}{c}-3 \\ 5 \\ -8\end{array}\right]\)

Step 5 ：\(\boxed{\operatorname{proj}_{V}(\vec{v}) = \left[\begin{array}{l}-3 \\ 5 \\ -8\end{array}\right]}\)