Find ABC

Multiply the three equations:

$a^2b^2c^2(a+b+c)^3=1001\cdot2002\cdot3003$

Now, multiply the three equations, two at time:

$ab^2c(a+b+c)^2=1001\cdot2002\\ a^2bc(a+b+c)^2=1001\cdot3003\\ abc^2(a+b+c)^2=2002\cdot3003$

Sum those three equations:

$abc(a+b+c)^3=1001\cdot2002+1001\cdot3003+2002\cdot3003$

Finally, divide the first equation we obtained by this one:

$abc= \dfrac{1001\cdot2002\cdot3003}{1001\cdot2002+1001\cdot3003+2002\cdot3003} \\ abc=\boxed{546}$

$1001/ab=2002/bc=3003/ca$ $1/ab=2/bc=3/ca$ $2a=c=3b$ $2a^2(a+2a+2/3a)=3003$ $22a^3=9009$ $abc=a*2a*(2/3)a=(4/3)a^3=(9009/22)*(4/3)=3003*2/11=6006/11=\boxed{546}$

lets take ratios, very simply, we can get that $ab:bc:ca=1:2:3$ now separate the ratios $ab:bc=1:2\longrightarrow a:c=1:2$ $ca:bc=3:2\longrightarrow a:b=3:2$ add these ratios to get $a:b:c=3:2:6$ and hence, $a=\dfrac{3}{2} b,c=3b,b=b$ so, $abc=\dfrac{9}{2}b^3$ now solve any 1 equation, $bc(a+b+c)=2002\longrightarrow 3b^2\times \dfrac{11}{2}b=2002$ and solve $b^3= \dfrac{364}{3}$ now insert it, $abc=\dfrac{9}{2}b^3=\dfrac{9}{2}\times\dfrac{364}{3}=\boxed{546}$

Dividing the first equation by the second and the first by the third, we obtain

$a=\frac{1}{2}c$ and $b=\frac{1}{3}c$

Substituting these values into the first equation, we obtain

$1001=ab(a+b+c)$

$\qquad=ab(\frac{1}{2}c+\frac{1}{3}c+c)$

$\qquad=\frac{11}{6}abc$

Hence, $abc=\boxed{546}$