Answer
(a) the set is orthogonal.
(b)
$$u_1=(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3})$$
$$v_1=(-\frac{\sqrt{2}}{2},0 ,\frac{\sqrt{2}}{2}),$$
the set $\{u_1,v_1\}$ is orthonormal set.
Work Step by Step
Let $u=(\sqrt{3},\sqrt{3},\sqrt{3})$ and $v=(-\sqrt{2},0,\sqrt{2})$, then we have
(a) since $$u\cdot v=-\sqrt{6}+\sqrt{6}=0,$$
then the set is orthogonal.
(b) to normalize the set, we have
$$u_1=\frac{1}{\|u\|}u=\frac{1}{3}(\sqrt{3},\sqrt{3},\sqrt{3})=(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3})$$
$$v_1=\frac{1}{\|v\|}v=\frac{1}{2}(-\sqrt{2},0,\sqrt{2})=(-\frac{\sqrt{2}}{2},0 ,\frac{\sqrt{2}}{2}).$$
Now, the set $\{u_1,v_1\}$ is orthonormal set.
