If $a = 24, b = 26, c = 28$ , then the value of $a^{2} + b^{2} + c^{2} - a b - b c - a c$ will be

0

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4

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12

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Solution

The correct option is B $12$
Let the value of $b = k$ ,
$∴ a = k - 2$ and $c = k + 2$
$a^{2} + b^{2} + c^{2} - a b - b c - c a = (a + b + c)^{2} - 3 (a b + b c + c a) = (3 k)^{2} - 3 (k (k + 2) + (k^{2} - 4) + k (k - 2)) = 9 k^{2} - 3 (3 k^{2} - 4) = 12$

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Similar questions

If a+b+c =0 , then the value of $\frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b}$

a + b + c = 9

a b + b c + c a = 26

a^{2} + b^{2} + c^{2} - 3

a + b + c = 9

a b + b c + a c = 26

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

\frac{a^{2} + b^{2} + a b}{b^{2} + c^{2} + b c} + \frac{c^{2} + c a + a^{2}}{b^{2} + c^{2} + b c}

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