Given, l+m+n=0 …… (I) 2lm+2ln−mn=0 …… (II) From equation (I) and (II), 2l(−l)−mn=0 2l2+mn=0 .......(III) Square both sides of equation (I), l2+m2+n2+2lm+2mn+2nl=0 As l2+m2+n2=1, So, 2lm+2mn+2nl=−1 .......(IV) From equation (II) and (IV), mn=−
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From equation (III), 6l2−1=0 So, l1l2=−
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From equation (I) and (II), 2lm+(2l−m)(−l−m)=0 2l2−lm−m2=0 2(