A) \[-1\]
B) 0
C) 1
D) 2
Correct Answer: C
Solution :
\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2a-2b-2c-3\] \[\Rightarrow \]\[{{a}^{2}}-2a+{{b}^{2}}+2b+{{c}^{2}}+2c+1+1+1=0\] \[\Rightarrow \]\[({{a}^{2}}-2a+1)+({{b}^{2}}+2b+1)+({{c}^{2}}+2c+1)=0\] \[\Rightarrow \]\[{{(a-1)}^{2}}+{{(b+1)}^{2}}+{{(c+1)}^{2}}=0\] \[\therefore \]\[a=1,\,\,b=-1\,\,and\,\,c=-1\] \[\therefore \]\[2a-3b+4c=2(+1)-3(-1)+4(-1)\] \[=2+3-4=5-4=1\]You need to login to perform this action.
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