It's given that, $ABC \times D = 37DD$

Wherein A, B, C and D are different non-zero digits.

So, $ABC = \frac{37DD}{D} = \frac{3700 + 10D + D}{D} = \frac{3700}{D} + 11$

The possible values of D, such that ABC is an integer, are 1, 2, 4, and 5.

If $D = 1$, $ABC = 3700 + 11 = 3711$. It can be rejected as ABC is a three-digit number. If $D = 2$, $ABC = 1850 + 11 = 1861$. It can be rejected as ABC is a three-digit number. If $D = 5$, $ABC = 740 + 11 = 751$. It can be rejected as here $B = D = 5$. If $D = 4$, $ABC = 925 + 11 = 936$.

So, $A + B + C = 9 + 3 + 6 = 18$

Hence, option A is correct.