In order for the number to be divisible by $11$, we must have $a+b+c+d \equiv 5 \mod 11$. For it to be divisible by $3$, we must have $a+b+c+d \equiv 1 \mod 3$. By CRT we can conclude that $a+b+c+d \equiv 16 \mod 33$. Since $a+b+c+d \leq 9+8+7+6 = 30$, we must have $a+b+c+d=16$. Clearly since $d$ is the greatest, we must have $d \geq 6$. $d=9$. Then $c+b+a=7$, which gives $4$ solutions. $d=8$. Then $c+b+a = 8$, giving $6$ solutions. $d=7$. Then $c+b+a = 9$, giving $5$ solutions. $d=6$. Then $c+b+a=10$, giving $2$ solutions. So unless I made some careless errors, there are in total, $17$ such numbers.

bzprules wrote: In order for the number to be divisible by $11$, we must have $a+b+c+d \equiv 5 \mod 11$. For it to be divisible by $3$, we must have $a+b+c+d \equiv 1 \mod 3$. By CRT we can conclude that $a+b+c+d \equiv 16 \mod 33$. Since $a+b+c+d \leq 9+8+7+6 = 30$, we must have $a+b+c+d=16$. Clearly since $d$ is the greatest, we must have $d \geq 6$. $d=9$. Then $c+b+a=7$, which gives $4$ solutions. $d=8$. Then $c+b+a = 8$, giving $6$ solutions. $d=7$. Then $c+b+a = 9$, giving $5$ solutions. $d=6$. Then $c+b+a=10$, giving $2$ solutions. So unless I made some careless errors, there are in total, $17$ such numbers. Are you sure about your answer? Did you remember all 4 digits are diferrent?

d=8 should give 5 solutions?