Let f x = a - 1 0 a x a - 1 a x 2 a x a , a ∈ R . Then the sum of the squares of all the values of a…

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

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

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

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

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

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

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

$\Rightarrow 2\times 2\mathrm{a}\left(10+\mathrm{a}\right)-2\mathrm{a}\left(5+\mathrm{a}\right)+100=0$

$\Rightarrow 40a+4a^2-10a-2a^2+100=0$

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

$\Rightarrow a^2+15a+50=0$

$\left(a+10\right)\left(a+5\right)=0$

$a=-10$ or $a=-5$

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