Do you know your alphabets?

Suppose we say the integers that makes it largest are $x<y<z$ in that order.

We now show that $y-x<3$ and $z-y<3$ .

The expression $xy+yz+zx$ is equal to $xy+(x+y)z$ .

If $y-x\geq 3$ , we can increase $x$ by one and decrease $y$ by 1.

$(x+1)(y-1)=xy+y-x-1\geq xy$

A similar argument follows for $z-y<3$ .

If $y-x=2$ and $z-y=2$ , we can increase $x$ by one and decrease $z$ by 1. A similar argument follows.

We consider mod 3. It is well known that if $x+y+z\equiv 0$ , then $x, y, z$ are congruent, or all different mod 3.

If one of $y-x, z-y$ is 2, then the other is 1. Regardless, $z-y+y-x=3$ , thus $x, z$ congruent mod 3. Thus $y$ is congruent to $x$ mod 3, but $0<y-x<3$ from above, and y cannot be congruent to x. Thus both of $y-x, z-y$ are 1.

Thus the only possible triplet is $(4, 5, 6)$ . Evaluating, we get 74.

The largest distinct numbers are 4, 5 and 6. $4 \times 5 + 5 \times 6 + 6 \times 4 = \boxed{74}$