If $a, b, c$ are positive integers such that $a > b > c$ 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

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

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$0 = 2.$

$\therefore a-b=1,b-c=1,c-a=-2\text{. }$

Now, $3a+7b-10c=3(a-c)+7(b-c)=13$