Partial answer (could be used as a springboard by somebody else)
Preliminary size considerations:
The right-hand side can't be larger than around $10,000$, so the left-hand side can't include $8!$ or $9!$, i.e. $A,C,D$ are all at most 7. But since $A,B,C,D$ are all positive and distinct, the right-hand side must be quite a bit bigger than $144$, so at least one of $A,C,D$ must be bigger than 5.
Case 1
If one of $A,C,D$ is 7, then
the left-hand side is between $5043$ and $5880$, so $DB$ must be seventy something, i.e. $D=7$.
[to be completed]
Case 2
If none of $A,C,D$ is 7, then
the left-hand side is at most $864$, so $DB$ is less than $30$, so $D$ is 1 or 2. We also know that one of $A,C$ must be 6.
Case 2a
If $A=6$, then the right-hand side modulo 4 is $B^2+B^2+B^2-3B=3B(B-1)$ which is even. All factorials after the first one are even, so we can't have $D=1$, which means $D=2$. So the possibilities for the left-hand side are $723,728,746,842$.
Considering each of these possibilities in turn:
$C=1,723=(6+B)^2+(2+B)^2+(20+B)^2-3B=3B^2+53B+440$, impossible.
$C=3,728=(6+B)^2+(6+B)^2+(20+B)^2-3B=3B^2+61B+472$, impossible.
$C=4,746=(6+B)^2+(8+B)^2+(20+B)^2-3B=3B^2+65B+500$, impossible.
$C=5,842=(6+B)^2+(10+B)^2+(20+B)^2-3B=3B^2+69B+536$, impossible.
So Case 2a is impossible.
Case 2b
If $C=6$, then ...
[to be completed]