Answer
See below
Work Step by Step
Assume $y_1=(1+i,i,2-i)\\
y_2=(1+2i,1-i,i)$
Apply Gram- Schmidt
$v_1=y_1=(1+i,i,2-i)\\
v_2=y_2-\frac{}{||v_1||^2}v_1\\
=(1+2i,1-i,i)-\frac{}{||(1+i,i,2-i)||^2}(1+i,i,2-i)\\
=(1+2i,1-i,i)-\frac{(1+2i)(\bar 1+i)+(1-i).\bar i+i(\bar 2-i)}{(1+i)(\bar 1+i)+i.\bar i+(2-i).(\bar 2-i)}(1+i,i,2-i)\\
=(1+2i,1-i,i)-\frac{(1+2i)(1-i)+(1-i).(-i)+i( 2+i)}{(1+i)(1-i)+i.(- i)+(2-i).(2+i)}(1+i,i,2-i)\\
=(1+2i,1-i,i)-\frac{1+2i}{1+1+1+4+1}(1+i,i,2-i)\\
=(1+2i,1-i,i)-\frac{1+2i}{8}(1+i,i,2-i)\\
=(1+2i,1-i,i)+\frac{1}{8}(1+i+2i+2i^2,i+2i^2,2-i+4i-2i^2)\\
=(1+2i,1-i,i)+\frac{1}{8}(-1+3i,-2+i,4+3i)\\
=\frac{1}{8}(13i+9,10-9i,5i-4)$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1+i,i,2-i)}{\sqrt (1+i)(\bar 1+i)+i.\bar i+(2-i)(\bar 2-i)}=\frac{(1+i,i,2-i)}{\sqrt (1+i)(1-i)+i.(-i)+(2-i)(2+i)}=\frac{(1+i,i,2-i)}{1-i^2-i^2+2^2-i^2}=\frac{(1+i,i,2-i)}{1+1+1+4+1}=\frac{1}{2\sqrt 2}(1+i,i,2-i)$
$\frac{v_2}{||v_2||}=\frac{\frac{1}{8}(13i+9,10-9i,5i-4)}{\sqrt \frac{1}{8}\frac{1}{8}(13i+9)(\bar 13i+9)+(10-9i)(\bar 10-9i)+(5i-4)(\bar 5i-4)}=\frac{\frac{1}{8}(13i+9,10-9i,5i-4)}{\frac{1}{8}\sqrt 9^2-(13i)^2+10^2-(9i)^2+(-4)^2-(5i)^2}=\frac{1}{2\sqrt 118}(13i+9,10-9i,5i-4)$
Consequently, an orthogonal basis for colspace $(A)$ is
$\{\frac{1}{2\sqrt 2}(1+i,i,2-i),\frac{1}{2\sqrt 118}(13i+9,10-9i,5i-4)\}$
