The smallest number

Since $a = 2b, 3b=c$ and $a+b+c=6b$
$6b=2b \times b \times 3b$
$=6b^3$
b=1 for the equation to be true.
$a=2b=2(1)=2$
$c=3b=3(1)=3$
$2\times1\times3=6$

Substituting a = 2b in a + b = c we get, c = 3b. Substituting a and c in a + b + c = abc we get, ac = 6. Now ac = 2b × 3b = 6b^2. Therefore comparing ac = 6 and ac = 6b^2 we get, b = 1 or -1. So the answer is abc = 6

Do we know that $1$ , $2$ and $3$ are the only set of numbers that satisfy $a+b+c=abc$ ? If that is the case, then the rest of the data can be ignored. I think so...but not sure. But all we know, is that $1$ , $2$ and $3$ are the $only$ $consecutive$ numbers to satisfy it...

If $a = 2b$ , then $(a+b=c) \rightarrow (3b=c)$ .
Then $(a+b+c=abc) \rightarrow (2b+b+3b=abc) \rightarrow (6b=abc) \rightarrow (6=ac)$ .
Then $(6=ac) \rightarrow (6=(2b)(3b)) \rightarrow (6=6b^2) \rightarrow (1=b^2) \rightarrow (b=1)$ .
Then $(a=2b) \rightarrow (a=2)$ .
Then $(c=3b) \rightarrow (c=3)$ .
Therefore, $abc=(2)(1)(3)=6$ .