Given the three vectors are
u1=[1,−3,−2]
u2=[5,9,−6]
u3=[5,−7,h]
Now the three vectors will be linearly independent when for three scalars, a,b and c , we can write,
au1+bu2+cu3=0 if and only iff when a=b=c=0
Therefore,
au1+bu2+cu3=a[1,−3,−2]+b[5,9,−6]+c[5,−7,h]=0
a[1,−3,−2]+b[5,9,−6]+c[5,−7,h]=[0,0,0]
[a,−3a,−2a]+[5b,9b,−6b]+[5c,−7c,hc]=[0,0,0]
a+5b+5c=0−3a+9b−7c=0−2a−6b+hc=0
Solving these equations,
since for the linearly independent, a=b=c=0 , which requires,
h+326=0
h=326
Hence other than this value, the three vectors will be linearly independent.
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