If a, b, c are positive integers such that a > b > c and $$111abca2b2c2=$$−$$2$$ then $$3a+7b$$−$$10c$$ equals

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If $a, b, c$ are positive integers such that $a > b > c$ and $|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array}| = - 2$ then $3 a + 7 b - 10 c$ equals

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Solution

We have $Δ = (a - b) (b - c) (c - a) = - 2$
As $a > b > c, a - b, b - c$ are positive integers and $c - a$ is a negative integers. Only possibilities are
$a - b = 2, b - c = 1, c - a = - 1$ (1)
or $a - b = 1, b - c = 2, c - a = - 1$ (2)
or $a - b = 1, b - c = 1, c - a = - 2$ (3)
(1) and (2) lead us to $0 = 2 .$
$∴ a - b = 1, b - c = 1, c - a = - 2 .$
Now, $3 a + 7 b - 10 c = 3 (a - c) + 7 (b - c) = 13$

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Introduction to Determinants

Remember concepts with our Masterclasses.

If a, b, c are positive integers such that a > b > c and 111abca2b2c2=−2 then 3a+7b−10c equals

Ready to Test Your Skills?

free study material

If $a, b, c$ are positive integers such that $a > b > c$ and $|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array}| = - 2$ then $3 a + 7 b - 10 c$ equals