Question : If $x^{2}+y^{2}+z^{2}=14$ and $xy+yz+zx=11$, then the value of $(x+y+z)^{2}$ is:
Option 1: 16
Option 2: 25
Option 3: 36
Option 4: 49
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Correct Answer: 36
Solution : Given: $x^{2}+y^{2}+z^{2}=14$ and $xy+yz+zx=11$ Thus, $(x+y+z)^{2}=x^2+y^2+z^2+2(xy+yz+zx)$ Putting the values, we get $(x+y+z)^{2}=14+2(11)$ ⇒ $(x+y+z)^{2}=14+22$ ⇒ $(x+y+z)^{2}=36$ Hence, the correct answer is 36.
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