The sum of all the 4 -digit distinct numbers that can be formed...

Digits are $1, 2, 2, 3$

total distinct numbers $\dfrac{4 !}{2 !} = 12 .$

total numbers when

$1$ at unit place is $3 .$
$2$ at unit place is $6$
$3$ at unit place is $3 .$

So, sum $= ( 3 + 12 + 9 ) \left({10}^{3} + {10}^{2} + 10 + 1\right)$

$= ( 1111 ) \times 24$

$= 26664$