A) 50
B) 52
C) 54
D) 55
Correct Answer: D
Solution :
\[\because \]\[x+y+z=15\] Squaring both sides, \[{{(x+y+z)}^{2}}={{(15)}^{2}}\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(xy+yz+zx)\] \[=225\] Putting the value of\[(xy+yz+zx)\]. \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\times 85=225\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+170=225\] \[\therefore \] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=225-170=55\]
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