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

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

1. A
2. B
3. C
4. D

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### Solution:

We have $\mathrm{\Delta }=\left(a-b\right)\left(b-c\right)\left(c-a\right)=-2$

As $a>b>c,a-b,b-c$ are positive integers and  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

Now, $3a+7b-10c=3\left(a-c\right)+7\left(b-c\right)=13$

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