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
The plane is $H = \{ x | \langle d, x \rangle = \alpha \}$, where $d=(1,1,-1)^T$, and $\alpha = -1$. Let $v = (1,1,2)^T$.
The nearest point on $H$ to $v$ can be found by solving $\langle d, v+\lambda d \rangle = \alpha$, which gives $\lambda = \frac{\alpha-\langle d, v \rangle}{\|d\|^2}$.
The shortest distance is given by $\|v-(v+\lambda d) \| = |\lambda| \|d \| = | \frac{\alpha-\langle d, v \rangle}{\|d\|} | = \frac{1}{\sqrt{3}}$.
Addendum: To see why this is basically the Gram Schmidt process (or part thereof):
Let $b_2,b_3$ be any two vectors such that $d,b_2,b_3$ span $\mathbb{R}^3$. Apply the Gram Schmidt process to get $n_1,n_2,n_3$. We have $n_1 = \frac{d}{\|d\|} $, of course, and since $n_2 \bot d$ and $n_3 \bot d$, we see that $n_2,n_3$ lie on the plane $\langle d , x \rangle = 0$. Furthermore, we see that the point $\frac{\alpha}{\|d\|} n_1$ lies on the plane $H$ above. Hence every point $p$ on the plane is give by $p=\frac{\alpha}{\|d\|} n_1 + \beta_2 n_2 + \beta_3 n_3$, where $\beta_i$ are arbitrary.
Now project $v$ onto $n_1,n_2,n_3$ to get $v = \gamma_1 n_1 + \gamma_2 n_2 + \gamma_3 n_3$. Then the distance (or rather distance squared) from $v$ to a point on the plane is given by $\|v-p\|^2 = |\gamma_1-\frac{\alpha}{\|d\|}|^2+ |\gamma_2-\beta_2|^2 + |\gamma_3-\beta_2|^2$, from which we see that the minimum distance (choosing $\beta_2, \beta_3$ appropriately) is $|\gamma_1-\frac{\alpha}{\|d\|}|$. Since $\gamma_1 = \langle v, n_1 \rangle = \langle v, \frac{d}{\|d\|} \rangle$, we obtain the value $| \frac{\alpha-\langle d, v \rangle}{\|d\|} |$ as above.
Since we do not need to compute the nearest point, we only need compute $\gamma_1$, hence we only need the first step of the Gram Schmidt process applied to $d$ followed by the projection of $v$ onto this first element.