Sum of Numbers on Cube Faces to Equal 2004

[anemone]

Positive integers are written on all the faces of a cube, one on each. At each corner (vertex) of the cube, the product of the numbers on the faces that meet at the corner is written. The sum of the numbers written at all the corners is 2004. If $T$ denotes the sum of the numbers on all the faces, find all the possible values of $T$.

 

[I like Serena]

\begin{tikzpicture}[
face/.pic = {
\node {#1};
\draw (-1,-1) rectangle (1, 1);
},
cube/.pic = {
\draw (0,0) pic {face=a};
\draw (1,-1) -- (2,-0.5) -- (2,1.5) -- (1,1) node at (1.5,0.25) {b};
\draw (2,1.5) -- (0,1.5) -- (-1,1) node at (0.5,1.25) {e};
}]
\draw (0,0) pic {face=a} (2,0) pic {face=b} (4,0) pic {face=c} (6,0) pic {face=d} (0,2) pic {face=e} (0,-2) pic {face=f};
\draw (9,0) pic {cube};
\end{tikzpicture}

Spoiler: My attempt

Let $a,b,c,d,e,f$ be the 6 faces where $(a,c)$, $(b,d)$, and $(e,f)$ are the pairs of opposing faces as shown in the drawing.
The sum of the corners is $(ab+bc+cd+da)e+(ab+bc+cd+da)f=(a(b+d) + c(b+d))(e+f) =(a+c)(b+d)(e+f)=2004$.

The list of suitable factorizations of $2004=2^2\cdot 3\cdot 167$ is $(3\cdot 4\cdot 167,\, 2\cdot 6\cdot 167,\,2\cdot 3\cdot 334,\,2\cdot 2\cdot 501)$.
We are looking for the possibilities for $T=a+b+c+d+e+f$, which is the sum of the 3 factors.
Those possibilities for $T$ are $174,\, 175,\, 339,\, 505$.