Answer
(a) the set is not orthogonal
(b) the set is not orthonormal
(c) the set is a basis for $R^3$
Work Step by Step
Let $u=(\frac{\sqrt 2}{3},0,-\frac{\sqrt 2}{6})$ and $v=(0,\frac{2\sqrt 5}{5},-\frac{\sqrt 5}{5}), w=(\frac{\sqrt 5}{5},0,\frac{ 1}{2})$, then we have
(a) since
$$u\cdot v=\frac{\sqrt{ 10}}{30}\neq0$$
then the set is not orthogonal.
(b) since the set is not orthogonal then it is not orthonormal.
(c) to check if it is a basis for $R^2$, consider the combination
$$a(\frac{\sqrt 2}{3},0,-\frac{\sqrt 2}{6})+b(0,\frac{2\sqrt 5}{5},-\frac{\sqrt 5}{5})+c(\frac{\sqrt 5}{5},0,\frac{ 1}{2})=(0,0,0)$$
then we have the system
$$\frac{\sqrt 2}{3}a+ \frac{\sqrt 5}{5}c=0, \quad \frac{2\sqrt 5}{5}b=0,\quad -\frac{\sqrt 2}{6}a-\frac{\sqrt 5}{5}b+\frac{ 1}{2}c=0.$$
It is clear that the above system has the solution $a=0, b=0,c=0$. Then the set is linearly independent and $R^3$ has dimension $3$, then it is a basis for $R^3$.
