We have the equation, $(a+1)(b+1)(c+1)=2abc$
$\Rightarrow ab+bc+ca+a+b+c+1=abc$
$\Rightarrow 2(a+b+c)=abc-ab-bc-ca+a+b+c-1$
$\Rightarrow 2(a+b+c)=(a-1)(b-1)(c-1)$
Now, substituting $x=a-1,y=b-1,z=c-1$, we get, \[ 2(x+y+z+3)=xyz......(1)\]
All the $x,y,z$ are natural number, because if any one of $a,b,c$ is 1, suppose its c, we'll have $(a+1)(b+1)=ab$, which is not possible.
WLOG assume that $x\geq y\geq z$,
If $x=1,x=y=z=1$ which doesnt satisfy $(1)$
If $x=2$ then the highest value of RHS is 8, while the LHS is always greater than 8.
So, $x\geq 3\Rightarrow \frac{3}{x} \leq1$
we have, $xyz=2x+2y+2z+6\leq 6x+6$
$\Rightarrow yz-6\leq \frac{6}{x}\leq 2$
$\Rightarrow yz\leq 8$
$\Rightarrow 8\geq yz\geq z^2$
So the value of $z$ is either 1 0r 2.
- Case 1: $z=1$
\[ 2(x+y+4)=xy\]
\[ (x-2)(y-2)=12\]
So, $(x-2,y-2)=(12,1),(6,2),(4,3)$
SO we get three solutions $(a,b,c)=(15,4,2),(9,5,2),(7,6,2)$
- Case 2: $z=2$
\[ 2(x+y+5)=2xy\]
\[ (x-1)(y-1)=6\]
So, $(x-1,y-1)=(6,1),(3,2)$
So, we get 2 more solutions $(a,b,c)=(8,3,3),(5,4,3)$
Therefore all the solutions to the diophantine equation is
$(a,b,c)=(15,4,2),(9,5,2),(7,6,2),(8,3,3),(5,4,3)$
And all of its permutations.$\blacksquare$