The set \begin{equation*} \set{\vect{x}_1,\,\vect{x}_2,\,\vect{x}_3} =\set{ \colvector{1\\2\\1},\, \colvector{-1\\0\\1},\, \colvector{2\\1\\1} } \end{equation*} is a linearly independent set, which the Gram-Schmidt Process (Theorem GSP) converts to an orthogonal set, and which can then be converted to the orthonormal set, \begin{equation*} B= \set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3} =\set{ \frac{1}{\sqrt{6}}\colvector{1\\2\\1},\, \frac{1}{\sqrt{2}}\colvector{-1\\0\\1},\, \frac{1}{\sqrt{3}}\colvector{1\\-1\\1} } \end{equation*} which is therefore an orthonormal basis of $\complex{3}$. With three vectors in $\complex{3}$, all with real number entries, the inner product (Definition IP) reduces to the usual “dot product” (or scalar product) and the orthogonal pairs of vectors can be interpreted as perpendicular pairs of directions. So the vectors in $B$ serve as replacements for our usual 3-D axes, or the usual 3-D unit vectors $\vec{i},\vec{j}$ and $\vec{k}$. We would like to decompose arbitrary vectors into “components” in the directions of each of these basis vectors. It is Theorem COB that tells us how to do this.
Suppose that we choose $\vect{w}=\colvector{2\\-1\\5}$. Compute \begin{align*} \innerproduct{\vect{v}_1}{\vect{w}}=\frac{5}{\sqrt{6}}&& \innerproduct{\vect{v}_2}{\vect{w}}=\frac{3}{\sqrt{2}}&& \innerproduct{\vect{v}_3}{\vect{w}}=\frac{8}{\sqrt{3}} \end{align*} then Theorem COB guarantees that \begin{equation*} \colvector{2\\-1\\5}= \frac{5}{\sqrt{6}}\left(\frac{1}{\sqrt{6}}\colvector{1\\2\\1}\right)+ \frac{3}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\colvector{-1\\0\\1}\right)+ \frac{8}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\colvector{1\\-1\\1}\right) \end{equation*} which you should be able to check easily, even if you do not have much patience.