That is against the rules!

Given that $a, b$ and $c$ are positive integers and $a<b<c$ ,

$\begin{aligned} \log(a+b+c) &= \log(a) + \log(b) + \log (c) \\ \log(a+b+c) &= \log(a\cdot b \cdot c) \\ \implies a+b+c &= a \cdot b \cdot c \\ \text{ suppose } a \geq 2, \text{ then } b\geq 3 \\ a+b+c &= a\cdot b \cdot c \\ \implies b+c &= a(bc - 1) \\ \text{min. of RHS } &= 2 ( 3c - 1) = 6c - 2 = c + (5c -2 ) \text{ > LHS as } b < 5c-2 \text{ for c } \geq 3 \implies \text{ contradiction } \implies \boxed{a = 1} \\ 1+b+c &= b\cdot c \\ \implies c(b-1) &= b+ 1 \implies c = \cfrac{b+1}{b-1} \implies \boxed{b = 2 } \implies 1+2+c=2c \implies \boxed{c = 3 }\\ \\ a^2 b^3 c^4 &= 1^2 2^3 3^4= 1\cdot 8 \cdot 81 = \boxed{648} \end{aligned}$

$\begin{aligned} \log(a+b+c) & = \log a + \log b + \log c \\ \log(a+b+c) & = \log(abc) \\ \implies a + b + c & = abc \end{aligned}$

This implies that $a=1$ , $b=2$ , and $c=3$ , and $a^2b^3c^4 = 1 \times 8 \times 81 = \boxed{648}$ .