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Question : 12 of 90
Marks:
+1,
-0
Solution:
[]=0‌‌∴||=0 ‌⇒b−1+c−1+a(1−bc)=0‌∴abc=a+b+c−2 []=0‌‌∴||=0‌⇒−2a(ac−bc−c2)+2c(a2+ab−ac)=0‌⇒−2a2c+2‌abc+2ac2+2a2c+2‌abc−2ac2=0‌⇒4abc=0‌‌∴abc=0‌∴a+b+c=2‌‌∴6(a+b+c)=12‌ Ans. ‌‌
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