If $a + b + c = 12$ and $a b + b c + c a = 60$ . Find the value of $a^{2} + b^{2} + c^{2}$ .

12

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24

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2

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- 12

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Solution

The correct option is B $24$
Given: $a + b + c = 12$
$a b + b c + c a = 60$
We know that square of trinomial identity, $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
Now, substitute the values, we get
$(12)^{2} = a^{2} + b^{2} + c^{2} + 2 (60)$
$144 - 120 = a^{2} + b^{2} + c^{2}$
$a^{2} + b^{2} + c^{2} = 24$

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a + b + c = 12

a^{2} + b^{2} + c^{2} = 50

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a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} [{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}]

a^{2} + b^{2} + c^{2} = 50

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a^{2} + b^{2} + c^{2} = 66

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