I’ve got you to $\displaystyle lcm(a_1,a_2,a_3)|gcd(a_2a_3+a_1a_3+a_1a_2 , a_1+a_2+a_3)$

Now, let’s keep going. Define $\displaystyle g=gcd(a_1,a_2,a_3)$ and let $\displaystyle u_1=a_1/g$ , $\displaystyle u_2=a_2/g$ , $\displaystyle u_3=a_3/g$ .

Then $\displaystyle u_1u_2u_3g|gcd((u_2u_3+u_1u_3+u_1u_2 )g^2, (u_1+u_2+u_3)g)$

Or, alternatively:

$\displaystyle u_1|u_2u_3(gk)+(u_2+u_3)$

$\displaystyle u_2|u_1u_3(gk)+(u_1+u_3)$

$\displaystyle u_3|u_2u_1(gk)+(u_2+u_1)$

In order to satisfy this for all $\displaystyle k$, $\displaystyle u_1u_2u_3|g$ and $\displaystyle u_1u_2u_3|u_1+u_2+u_3$ . This calls for three relatively prime integers whose product divides their sum. I believe $\displaystyle 1,2,3$ is the only such arrangement, leading us into a dead end.

However, after looking at some raw data, it looks like $\displaystyle gcd(p_1-1,p_2-1,p_3-1)$ and subsequent $\displaystyle p_1/g, p_2/g, p_3/g$ are all over the map. This leads me to pose another conjecture:

Weak: For any three relatively prime $\displaystyle u_1<u_2<u_3$, there exists a $\displaystyle k$ such that $\displaystyle (u_1k+1) (u_2k+1) (u_3k+1)$ is a three-factor Carmichael Number.

Strong: For any three relatively prime $\displaystyle u_1<u_2<u_3$, there exists an $\displaystyle a,b$ such that $\displaystyle (u_1k+1) (u_2k+1) (u_3k+1)$ is a three-factor Carmichael Number iff $\displaystyle k \equiv b (\bmod a)$ and $\displaystyle u_ik+1$ is prime for all $\displaystyle i$.

Examples:

For $\displaystyle u=(1,2,3), a=6, b=0$ generates Carmichaels 1729, 294409, 56052361, 118901521 …

For $\displaystyle u=(1,3,5), a=30, b=12$ generates Carmichaels 29341, 1152271, 64377911, 775368901 …

For $\displaystyle u=(1,4,7), a=28, b=4$ generates Carmichaels 2465, 6189121, 19384289, 34153717249 …