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Question: Answered & Verified by Expert
Let a,b,cR+ and the system of equations (1-a)x+y+z=0,x+(1-b)y+z=0 and x+y+(1-c)z=0 has infinitely many solutions, the minimum value of abc is
MathematicsDeterminantsJEE Main
Options:
  • A 33
  • B 9
  • C 27
  • D 3
Solution:
2022 Upvotes Verified Answer
The correct answer is: 27

We have,

1-ax+y+z=0

x+1-by+z=0

x+y+1-cz=0

has many solutions 

 1-a1111-b1111-c=0

 ab+ac+bc=abc

 ab+bc+ac3a2b2c21/3abc1/33

 Minimum of abc=33=27

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