Solution: We wil only check the set $\{ \mathbf{v} _1, \mathbf{v} _2, \mathbf{v} _3 \} $ are mutually orthogonal. Since orthogonal vectors are linearly independent, these vectors will be linearly independent. As the set $\{ \mathbf{v} _1, \mathbf{v} _2, \mathbf{v} _3 \} $ contains three linearly independent vectors, it will be a basis of $\mathbb{R} ^3$. Note that \begin{align*} \mathbf{v} _1 \cdot \mathbf{v} _2 & = -1 + 0 + 1 = 0 \\ \mathbf{v} _2 \cdot \mathbf{v} _3 & = -2 + 4 -2 = 0 \\ \mathbf{v} _3 \cdot \mathbf{v} _1 & = 2 + 0 -2 = 0 . \end{align*} Thus, $\{ \mathbf{v} _1, \mathbf{v} _2, \mathbf{v} _3 \} $ is an orthogonal basis of $\mathbb{R} ^3$.

Now we will normalize $\mathbf{v} _i$ to obtain an orthonormal basis of $\mathbb{R} ^3$. \begin{align*} \mathbf{u} _1 & = \frac{\mathbf{v} _1}{\lVert \mathbf{v} _1 \rVert } = \begin{bmatrix} \frac{1}{\sqrt{2} } \\[2ex] 0 \\ \frac{1}{\sqrt{2} } \\[2ex] \end{bmatrix} \\[2ex] \mathbf{u} _2 & = \frac{\mathbf{v} _2}{\lVert \mathbf{v} _2 \rVert } = \begin{bmatrix} -\frac{1}{\sqrt{18} } \\[2ex] \frac{4}{\sqrt{18} } \\[2ex] \frac{1}{\sqrt{18} } \\ \end{bmatrix} \\[2ex] \mathbf{u} _3 & = \frac{\mathbf{v} _3}{\lVert \mathbf{v} _3 \rVert } = \begin{bmatrix} \frac{2}{3} \\[1ex] \frac{1}{3} \\[1ex] -\frac{2}{3} \\ \end{bmatrix}. \end{align*} Thus, $\mathcal{B} = \{ \mathbf{u} _1, \mathbf{u} _2, \mathbf{u} _3 \} $ is an orthonormal basis for $\mathbb{R} ^3$.

Finally, computing the coordinates with respect to $\mathcal{B} $ is clear. We just need to take the inner product with each $\mathbf{u} _i$'s. Thus, \[ [\mathbf{x} ]_{\mathcal{B} } = \begin{bmatrix} \mathbf{u} _1 \cdot \mathbf{x} \\ \mathbf{u} _2 \cdot \mathbf{x} \\ \mathbf{u} _3 \cdot \mathbf{x} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{2}{\sqrt{18} } \\[2ex] \frac{5}{3} \\ \end{bmatrix}. \]