If $a + b + c = 6$ and $a b + b c + c a = 11$
Find $(a^{2} + b^{2} + c^{2})$ ?

14

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25

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36

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47

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Solution

The correct option is A $14$
Given $a + b + c = 6$

$(a + b + c)^{2} = 36$

$(a^{2} + b^{2} + c^{2}) + 2 (a b + b a + c a) = 36$

$a^{2} + b^{2} + c^{2} + (2 * 11) = 36$

$a^{2} + b^{2} + c^{2} = 36 - 22 = 14$

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