Let f(x) = | cases a & - 1 & 0 ax & a & - 1 a x ^ 2 & ax & a cases...

Given, f\left(x\right) = \left|\begin{cases}a & - 1 & 0 \\ ax & a & - 1 \\ a{x}^{2} & ax & a\end{cases}\right|

f\left(x\right) = a\left|\begin{cases}1 & - 1 & 0 \\ x & a & - 1 \\ {x}^{2} & ax & a\end{cases}\right|

$= a\left[1\left({a}^{2} + ax\right) + 1\left(ax + {x}^{2}\right)\right]$

$⇒ f\left(x\right) = a{\left(x + a\right)}^{2}$

Now differentiating the function w.r.t $x$

We get, ${f}^{'}\left(x\right) = 2a\left(x + a\right)$

Given, $2{f}^{'}\left(10\right) - {f}^{'}\left(5\right) + 100 = 0$

$⇒ 2 \times 2a\left(10 + a\right) - 2a\left(5 + a\right) + 100 = 0$

$⇒ 40a + 4{a}^{2} - 10a - 2{a}^{2} + 100 = 0$

$2{a}^{2} + 30a + 100 = 0$

$⇒ {a}^{2} + 15a + 50 = 0$

\left(a + 10\right)\left(a + 5\right) = 0$</p><p>$a = - 10 or $a = - 5$

Required $= {\left(- 10\right)}^{2} + {\left(- 5\right)}^{2} = 125$