l+m+n=0−(i)l2+m2+n2=1−(ii)l2+m2−n2=0−(iii)(ii)+(iii)2(l2+m2)=12n2=1⇒n=±1√2l2+m2=12l+m=1√2l2+(1√2−l)2=12l2+l2+12−√2l=122l2−√2l=0l=0,1√2⇒m=1√2,0l2+m2=12l+m=−1√2l2+(−1√2−l)2=12l2+12+l2+√2l=122l2+√2l=0l=0,−1√2⇒−1√2,0
Possible values of (l,m,n) are:
(0,1√2,−1√2),(1√2,0,−1√2),(0,−1√2,1√2),(−1√2,0,1√2)