(3)

We have,

$f\left(x\right)=\frac{x}{3}+\frac{3}{x}+3$

$\Rightarrow f\text{'}\left(x\right)=\frac{1}{3}–\frac{3}{x^2}$

For critical points,

$f\text{'}\left(x\right)=0 \Rightarrow \frac{1}{3}–\frac{3}{x^2}=0$

$\Rightarrow x^2=9 \Rightarrow x=\pm 3$

Now, $f\text{'}\left(x\right) >0\forall x\in (–\infty ,–3)\cup (3,\infty )$

and $f\text{'}\left(x\right)<0\forall x\in (–3,0)\cup (0,3)$

$\therefore$   f(x) is increasing in $(–\infty ,–3)\cup (3,\infty )$ and decreasing in $(–3,0)\cup (0,3)$

$\therefore {\alpha }_1=–3, {\alpha }_2=3, {\alpha }_3=–3, {\alpha }_4=0, {\alpha }_5=3$

$\therefore \sum _{i=1}^5{\alpha }_i^2=9+9+9+0+9=36$.

(1)

Given : $f\left(x\right)=2 {\log }_{e }(x–2)–x^2+ax+1$ is strictly increasing on (2, 3)

$g\left(x\right)={(x–1)}^3{(x+2–a)}^2$ is strictly decreasing on (b, c).

Using f(x), 

         $f\text{'}\left(x\right)=\frac{2}{x–2}–2x+a$

$\Rightarrow f\text{'}\text{'}\left(x\right)=\frac{–2}{{(x–2)}^2}–2<0$

SInce f(x) is strictly increasing, so $f\text{'}\left(x\right)>0$.

But we have given that (2, 3) is the largest open interval where f(x) is strictly increasing.

$\Rightarrow f\text{'}\left(3\right)=0 \Rightarrow 2–6+a=0 \Rightarrow a=4$

Taking, $g\left(x\right)={(x–1)}^3{(x+2–a)}^2$

                         $={(x–1)}^3{(x–2)}^2$           [$\because$ a = 4]

$\Rightarrow g\text{'}\left(x\right)={(x–1)}^2(x–2)(2x–2+3x–6)$

                  $={(x–1)}^2(x–2)(5x–8)$

Here, $g\text{'}\left(x\right)<0$          [$\because$ g(x) is strictly decreasing]

$\Rightarrow {(x–1)}^2(x–2)(5x–8)<0$

$\Rightarrow x\in (\frac{8}{5},2)$

$\Rightarrow b=\frac{8}{5} \mathrm{and} c=2$

Finally, we get 100 (a + b – c) = $100(4+\frac{8}{5}–2)=360$.