Your $u_1$ is correct, but your $u_2$ is incorrect; as noted in the comments, $u_1\cdot u_2 \neq 0$.
Recall that the standard inner product on $\mathbb{C}^n$ is given by $\langle u, v\rangle = u\cdot\bar{v}$. With this in mind, let's calculate $u_2$. First we have
$$\langle v_2, u_1\rangle = (-1, i, 1)\cdot\overline{\frac{1}{\sqrt{2}}(1, 0, i)} = \frac{1}{\sqrt{2}}(-1, i, 1)\cdot(1, 0, -i) = \frac{-1-i}{\sqrt{2}},$$
so
$$w_2 = v_2 - \langle v_2, u_1\rangle u_1 = \left[\begin{array}{c} -1\\ i\\ 1\end{array}\right] + \frac{1+i}{\sqrt{2}}\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0\\ i\end{array}\right] = \left[\begin{array}{c} \frac{-1+i}{2}\\ i\\ \frac{1+i}{2}\end{array}\right].$$
As
\begin{align*}
\|w_2\|^2 &= \sqrt{\left|\frac{-1+i}{2}\right|^2 + |i|^2 + \left|\frac{1+i}{2}\right|^2}\\
&= \sqrt{\left(\frac{-1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 0^2 + 1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2}\\
&= \sqrt{2}
\end{align*}
we have
$$u_2 = \frac{1}{\|w_2\|}w_2 = \frac{1}{\sqrt{2}}\left[\begin{array}{c} \frac{-1+i}{2}\\ i\\ \frac{1+i}{2}\end{array}\right].$$
Let's check to see if $u_2$ is orthogonal to $u_1$:
\begin{align*}
\langle u_1, u_2\rangle &= \frac{1}{\sqrt{2}}(1, 0, i)\cdot\overline{\frac{1}{\sqrt{2}}\left(\frac{-1+i}{2}, i, \frac{1+i}{2}\right)}\\
&= \frac{1}{2}(1, 0, i)\cdot\left(\frac{-1-i}{2}, -i, \frac{1-i}{2}\right)\\
&= \frac{1}{2}\left(\frac{-1-i}{2} + 0 + \frac{1+i}{2}\right)\\
&= 0.
\end{align*}
Note, we didn't have to normalise before we checked orthogonality; i.e. we could have checked $\langle u_1, w_2\rangle = 0$ instead.
I won't do the calculation of $u_3$ now. It is similar, but there are more computations. Note however that you have a typo in your formula for $w_3$; it should be
$$w_3 = v_3 - \langle v_3, u_1\rangle u_1 - \langle v_3, u_2\rangle u_2.$$
Now that you have the correct $u_2$ and the correct formula for $w_3$, the computation for $u_3$ should work out and produce $u_3 = \frac{1}{2}(i, -1-i, 1)$.
Best Answer
You made a small mistake, it should be:
$$u_2-\langle u_2,V_1\rangle V_1=\left<\frac{11}{14},\frac{-9}{14},\frac{\color{red}{8}}{14}\right>^T$$
and then
$$V_2 = \left<\frac{11}{\sqrt{266}},\frac{-9}{\sqrt{266}},\frac{8}{\sqrt{266}}\right>^T$$
Now you can check that $\|V_1\| = \|V_2\| = 1$ and $\langle V_1, V_2\rangle = 0$ so they are correct.