If $a_1,a_2\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{a}_3$ are the three values of a which satisfy the equation $\underset{0}{\overset{\pi /2}{\int }}{(\sin x+a\cos x)}^{3}dx-\frac{4a}{\pi -2}\underset{0}{\overset{\pi /2}{\int }}x\cos xdx=2$  then find the value of  $(a_1^2+a_2^2+a_3^2).$

  $\underset{0}{\overset{\pi /2}{\int }}{(\sin x+\alpha \cos x)}^{3}dx-\frac{4\alpha }{\pi -2}\underset{0}{\overset{\pi /2}{\int }}x\cos xdx=2$
Let  $I_1=\underset{0}{\overset{\pi /2}{\int }}{(\sin x+\alpha \cos x)}^{3}dx$
 $=\underset{0}{\overset{\pi /2}{\int }}{({\sin }^{3}x+{\alpha }^3{\cos }^{3}x+3\alpha {\sin }^{2}x\cos x+3{\alpha }^2\sin x{\cos }^{2}x)}^{3}dx$
 
 $=\underset{0}{\overset{\pi /2}{\int }}{\sin }^{3}xdx+{\alpha }^3\underset{0}{\overset{\pi /2}{\int }}{\cos }^{3}xdx+3\alpha \underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}x\cos xdx+3{\alpha }^2\underset{0}{\overset{\pi /2}{\int }}\sin x{\cos }^{2}xdx$$=\frac{2}{3}+{\alpha }^3\left(\frac{2}{3}\right)+3\alpha \underset{0}{\overset{1}{\int }}{t}^{2}dt+3{\alpha }^2\underset{0}{\overset{1}{\int }}{t}^{2}dt$
 $(\sin x=t\Rightarrow \cos xdx=dt;\underset{0}{\overset{1}{\int }}{t}^{2}dt)$
 
 $=\frac{2}{3}(1+{\alpha }^3)+\alpha +{\alpha }^2=\frac{2}{3}+\frac{2{\alpha }^3}{3}+\alpha +{\alpha }^2$
 
 $I_1=\frac{2{\alpha }^3}{3}+{\alpha }^2+\alpha +\frac{2}{3}$
 
 $I_2=\underset{0}{\overset{\pi /2}{\int }}\underset{I}{\underbrace{x}}\cdot \underset{II}{\underbrace{\cos x}}dx={\left[x\sin x\right]}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}\sin xdx$
 $I_2=\frac{\pi }{2}-1=\frac{\pi -2}{2}$
 
 $I=\frac{2{\alpha }^3}{3}+{\alpha }^2+\alpha +\frac{2}{3}-\frac{4\alpha }{\pi -2}-\frac{\pi -2}{2}$
 
 $\Rightarrow \frac{2{\alpha }^3}{3}+{\alpha }^2-\alpha +\frac{2}{3}=2$
 $\Rightarrow 2{\alpha }^3+3{\alpha }^3-3\alpha +2=6$
$\Rightarrow 2{\alpha }^3+3{\alpha }^3-3\alpha -4=0$

${\alpha }_1+{\alpha }_2+{\alpha }_3=\frac{3}{2}$

$\Rightarrow \sum {\alpha }_1{\alpha }_2=-\frac{3}{2}$

$\therefore \sum a_1^2=\frac{9}{4}+\frac{6}{2}=\frac{21}{4}$

$\therefore 1000\sum a_1^2=1000\times \frac{21}{4}=250\times 21=5250$