Step 1:
Let the sides of $\Delta ABC$ be $x,x+1$ and $x+2$ where $x > 0$ and it is a natural number.
Let $a=x+2,b=x+1,c=x$
Since $x>0$
Therefore $a > b> c\Rightarrow A > B > C$
Let the smallest angle be $\theta$ then $C=\theta$ and greatest angle $A=2\theta$
Applying sine rule in $\Delta$le ABC we get,
$\large\frac{x}{\sin\theta}=\frac{x+2}{\sin 2\theta}=\frac{x+1}{\sin (180^{\large\circ}-3\theta)}$
Step 2:
Let consider first and second terms
$\large\frac{x}{\sin\theta}=\frac{x+2}{\sin 2\theta}$
$\sin 2\theta=2\sin\theta\cos\theta$
$\large\frac{x}{\cos\theta}=\frac{x+2}{2\sin\theta\cos\theta}$
$\Rightarrow 2\cos\theta=\large\frac{x+2}{x}$
Step 3:
Let consider first and third terms we get
$\large\frac{x}{\sin\theta}=\frac{x+1}{\sin 3\theta}$
$\large\frac{x}{\sin\theta}=\frac{x+1}{3\sin\theta-4\sin^3\theta}$
$3-4\sin^2\theta=\large\frac{x+1}{x}$
$3-4(1-\cos^2\theta)=\large\frac{x+1}{x}$
$4\cos^2\theta=\large\frac{x+1}{x}$$+1$
$\Rightarrow \large\frac{2x+1}{x}$
$\big(\large\frac{{x+2}^2}{x}\big)^2=\frac{2x+1}{x}$
Step 4:
On simplifying we get,
$\large\frac{1}{x}\big[\large\frac{x^2+4x+4}{x}$$-(2x+1)\big]=0$
$\large\frac{x^2+4x+4-2x^2-x}{x}$$=0$
$x^2+3x+4=0$
Or $x^2-3x-4=0$
$(x-4)(x+1)=0$
$\Rightarrow x=4$ or $x=-1$
$\Rightarrow x=4$
Hence the sides of $\Delta$le are $4,5,6$
Hence (a) is the correct answer.