So $W=$ span$\left\{\begin{bmatrix}i\\1\\0\end{bmatrix},\begin{bmatrix}-1\\1+i\\1\end{bmatrix}\right\}$ is a subspace under complex space $\Bbb{C}^3$. To get an orthogonal basis, we use the Gram-schmidt process, in particular $\langle x,y\rangle=x_1\bar{y}_1+...+x_n\bar{y}_n$.
$$\eqalign{v_1&=\begin{bmatrix}i\\1\\0\end{bmatrix}\cr v_2&=\begin{bmatrix}-1\\1+i\\1\end{bmatrix}-\frac{\left\langle\begin{bmatrix}-1\\1+i\\1\end{bmatrix},\begin{bmatrix}i\\1\\0\end{bmatrix}\right\rangle}{\left\langle\begin{bmatrix}i\\1\\0\end{bmatrix},\begin{bmatrix}i\\1\\0\end{bmatrix}\right\rangle} \begin{bmatrix}i\\1\\0\end{bmatrix} \cr&=\begin{bmatrix}-1\\1+i\\1\end{bmatrix}-\frac{\begin{bmatrix}-1\\1+i\\1\end{bmatrix}\begin{bmatrix}-i\\1\\0\end{bmatrix}}{\begin{bmatrix}i\\1\\0\end{bmatrix}\begin{bmatrix}-i\\1\\0\end{bmatrix}} \begin{bmatrix}i\\1\\0\end{bmatrix} =\begin{bmatrix}-1\\1+i\\1\end{bmatrix}-\frac{1+2i}{2} \,\begin{bmatrix}i\\1\\0\end{bmatrix}\cr &=\frac{1}{2}\,\begin{bmatrix}-2\\2+2i\\2\end{bmatrix}-\frac{1}{2} \,\begin{bmatrix}i-2\\1+2i\\0\end{bmatrix} =\frac{1}{2}\,\begin{bmatrix}-2\\2+2i\\2\end{bmatrix}-\frac{1}{2} \,\begin{bmatrix}i-2\\1+2i\\0\end{bmatrix} =\frac12 \,\begin{bmatrix}-i\\1\\2\end{bmatrix}\cr }$$
So we have $\begin{bmatrix}-i\\1\\2\end{bmatrix}$ and it works. But why is the third slot $2$ but not other numbers?
In addition to having $v_1 \cdot v_2 = 0$, you must also have ${\rm span}(v_1,v_2)=W$. If you change the last $2$, then you change the span.