Answer:
Let 
Applying \[{{C}_{3}}\to {{C}_{3}}+{{C}_{2}},\] we get 
Taking \[(x+y+z)\]common from \[{{C}_{3}},\]we get 
Here, two columns \[{{C}_{1}}\] and \[{{C}_{3}}\] are identical. \[\therefore \] \[A=0\]
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