The number of 3-digit odd numbers, whose sum of digits is a multiple of $$7$$, is ______.

Let the 3-digit odd number be

$$100a+10b+c$$

where

$$a\in\{1,2,\ldots,9\},\qquad b\in\{0,1,\ldots,9\}$$

and since the number is odd,

$$c\in\{1,3,5,7,9\}$$

Also,

$$a+b+c$$

must be a multiple of $$7$$.

Possible multiples of $$7$$ between minimum sum $$2$$ and maximum sum $$27$$ are

$$7,\ 14,\ 21$$

Now, count case-wise.

Case 1: $$c=1$$

Then,

$$a+b=6,13,20$$

For

$$a+b=6$$

number of solutions $$=6$$

For

$$a+b=13$$

number of solutions $$=6$$

For

$$a+b=20$$

number of solutions $$=0$$

Total cases $$=12$$

Case 2: $$c=3$$

Then,

$$a+b=4,11,18$$

For

$$a+b=4$$

number of solutions $$=4$$

For

$$a+b=11$$

number of solutions $$=8$$

For

$$a+b=18$$

number of solutions $$=1$$

Total cases $$=13$$

Case 3: $$c=5$$

Then,

$$a+b=2,9,16$$

For

$$a+b=2$$

number of solutions $$=2$$

For

$$a+b=9$$

number of solutions $$=9$$

For

$$a+b=16$$

number of solutions $$=3$$

Total cases $$=14$$

Case 4: $$c=7$$

Then,

$$a+b=0,7,14,21$$

For

$$a+b=0$$

number of solutions $$=0$$

For

$$a+b=7$$

number of solutions $$=7$$

For

$$a+b=14$$

number of solutions $$=5$$

For

$$a+b=21$$

number of solutions $$=0$$

Total cases $$=12$$

Case 5: $$c=9$$

Then,

$$a+b=5,12,19$$

For

$$a+b=5$$

number of solutions $$=5$$

For

$$a+b=12$$

number of solutions $$=7$$

For

$$a+b=19$$

number of solutions $$=0$$

Total cases $$=12$$

Hence, total number of required numbers

$$=12+13+14+12+12$$

$$=63$$

Therefore, the required number is $$\boxed{63}$$.