(C)

$2b=a+ c$                     $[\because a,b,c$ are in A.P.$]$                   ...(i)

$b^2=(a+1)(c+3)$        $[\because a+1,b,c+3$ are in G.P.$]$     ...(ii)

Also, $\frac{a+b+c}{3}=8$        $[\because$ A.M. of $a,b,c]$                       ...(iii)

$\Rightarrow \frac{3b}{3}=8 \Rightarrow b=8$                                                     [Using (i)]

By (ii), $64=(a+1)(c+3)$ and by (iii), $c=16-a$

$\Rightarrow 64=(a+1)(16-a+3)$

$\Rightarrow a^2-18a+45=0 \Rightarrow a=15,3$

$\therefore$   $a=15$                                                       $(a\ne 3,\text{as }a>10)$

Also, $c=1$

$\therefore$   ${\left[{(a·b·c)}^{1/3}\right]}^3=15\times 8\times 1=120$

(D)

$f\left(x\right)=\frac{1}{2+\sin 3x+\cos 3x}$

Now, $-\sqrt{2}\le \sin 3x+\cos 3x\le \sqrt{2}$

$\frac{1}{2+\sqrt{2}}\le \frac{1}{2+\sin 3x+\cos 3x}\le \frac{1}{2-\sqrt{2}}$

$[a,b]=[\frac{{\displaystyle 1}}{{\displaystyle 2+\sqrt{2}}},\frac{{\displaystyle 1}}{{\displaystyle 2-\sqrt{2}}}]$

Now, A.M. of $a$ and $b$ i.e., $\alpha =\frac{1}{2}[\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}]$

$\frac{1}{2}\left[\frac{4}{4-2}\right]=1$

$\beta =\text{G.M. of }a\text{ and }b=\sqrt{\frac{1}{2+\sqrt{2}}·\frac{1}{2-\sqrt{2}}}=\sqrt{\frac{1}{4-2}}=\frac{1}{\sqrt{2}}$

So, $\frac{\alpha }{\beta }=\frac{{\displaystyle \frac{1}{1}}}{\sqrt{2}}=\sqrt{2}$