If $(a + b + c) = 14$ and $(a^{2} + b^{2} + c^{2}) = 74$ , find the value of $(a c + b c + c a)$ .

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Solution

Given :- $(a + b + c) = 14$
$a^{2} + b^{2} + c^{2} = 74$
To find $a b + b c + c a$
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
$(14)^{2} = 74 + 2 (a b + b c + c a)$
$196 - 74 = 2 (a b + c + c a)$
$122 = 2 (a b + b c + c a)$
$(a b + b c + c a) = \frac{122}{2} = 61$
$∴ (a b + b c + c a) = 61$

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