Answer
See below
Work Step by Step
Assume $y_1=(1-i,0,i)\\
y_2=(1,1+i,0)$
Apply Gram- Schmidt
$v_1=y_1=(1+i,i,2-i)\\
v_2=y_2-\frac{}{||v_1||^2}v_1\\
=(1,1+i,0)-\frac{}{||(1+i,i,2-i)||^2}(1-i,0,i)\\
=(1,1+i,0)-\frac{1(\bar 1-i)+(1+i).0+0.(\bar i)}{(1-i)(\bar 1+i)+0.0+i.(\bar i)}(1-i,0,i)\\
=(1,1+i,0)-\frac{1+i}{(1+i)(1-i)+i.(- i)}(1-i,0,i)\\
=(1,1+i,0)-\frac{1+i}{1^2-i^2-i^2}(1-i,0,i)\\
=(1,1+i,0)-\frac{1+i}{3}(1-i,0,i)\\
=(1,1+i,0)+\frac{1}{3}(1-i^2,0,1-i)\\
=(1,1+i,0)+\frac{1}{3}(2,0,1-i)\\
=\frac{1}{3}(1,3+3i,1-i)$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1-i,0,i)}{\sqrt (1-i)(\bar 1-i)+0.0+i(\bar i)}=\frac{(1-i,0,i)}{\sqrt (1+i)(1-i)+i.(-i)}=\frac{(1-i,0,i)}{1-i^2-i^2}=\frac{(1-i,0,i)}{1+1+1}=\frac{1}{\sqrt 3}(1-i,0,i)$
$\frac{v_2}{||v_2||}=\frac{\frac{1}{3}(1,3i+3,1-i)}{\sqrt \frac{1}{3} \frac{1}{1.1+(3i+3)(\bar 3-3i)+(1-i)(\bar 1+i)}}=\frac{\frac{1}{3}(1,3i+3,1-i)}{\frac{1}{3}\sqrt 1+3^2-(3i)^2+1^2-i^2}=\frac{1}{\sqrt 21}(1,3i+3,1-i)$
Consequently, an orthogonal basis for colspace $(A)$ is
$\{\frac{1}{\sqrt 3}(1-i,0,i),\frac{1}{\sqrt 21}(1,3i+3,1-i)\}$
