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$a^{2} + b^{2} + c^{2} - a b - b c - c a =$

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

{(a - b - c)}^{2}

{(a - b + c)}^{2}

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

answer is D.

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Detailed Solution

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

$\Rightarrow \frac{1}{2} [a^{2} + b^{2} - 2 a b + b^{2} + c^{2} - 2 b c + c^{2} + a^{2} - 2 c a]$

$\Rightarrow \frac{1}{2} [2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - b c - 2 c a]$

$\Rightarrow \frac{1}{2} [2 c a^{2} + b^{2} + c^{2} - a b - b c - c a]$

$\Rightarrow a^{2} + b^{2} + c^{2} - a b - b c - c a$

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

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