A) 3
B) 4
C) 5
D) 7
Correct Answer: D
Solution :
\[{{x}^{2}}+{{y}^{2}}+\text{ }{{z}^{2}}=2\left( x+y+z \right)-3\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2x-2y-2z+3=0\] \[\Rightarrow {{x}^{2}}-2x+1+{{y}^{2}}-2y+1+{{z}^{2}}-2z+1=0\] \[\Rightarrow {{(x-1)}^{2}}+{{(y-1)}^{2}}+{{(z-1)}^{2}}=0\] \[\therefore \]\[x-1=0\Rightarrow x=1\] \[y-1=0\Rightarrow y=1\] \[z-1=0\Rightarrow z=1\] \[\therefore \,\,2x\,+2y+3z=2+2+3=7\]You need to login to perform this action.
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