Answer
$$[x]_B=\left[ \frac{5\sqrt{10}}{10} \quad -2 \quad \frac{5\sqrt{10}}{10}\right]^T.$$
Work Step by Step
Let $x=(2,-2,1)$ and $B=\{ (\frac{\sqrt{10}}{10},0,\frac{3\sqrt{10}}{10}),(0,1,0),(\frac{-3\sqrt{10}}{10},0,\frac{\sqrt{10}}{10}) \}$
To find the coordinates of $x$ relative to $B$, we have to find the following;
$$\langle (2,-2,1),(\frac{\sqrt{10}}{10},0,\frac{3\sqrt{10}}{10})\rangle =\frac{2\sqrt{10}}{10}+\frac{3\sqrt{10}}{10}=\frac{5\sqrt{10}}{10} $$
$$\langle(2,-2,1),(0,1,0)\rangle =-2$$
$$\langle(2,-2,1),(\frac{-3\sqrt{10}}{10},0,\frac{\sqrt{10}}{10})\rangle =\frac{-6\sqrt{10}}{10}+\frac{\sqrt{10}}{10} =-\frac{5\sqrt{10}}{10}.$$
Then, the coordinate matrix of $x$ relative to $B$ is given by
$$[x]_B=\left[ \frac{5\sqrt{10}}{10} \quad -2 \quad \frac{5\sqrt{10}}{10}\right]^T.$$
