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

$$\begin{gathered} \text{ x y }z\leftarrow \text{ odd number } \\ z=1,3,5,7,9 \\ x+y+z=7,14,21 \text{ [sum of digit multiple of }7] \\ \underset{1\text{ to9 }}{\mathrm{x}}+\underset{\text{ 0 to 9 }}{\mathrm{y}}=6,4,2,13,11,9,7,5,20,18,16,14,12 \\ x+y=6\Rightarrow (1,5),(2,4),(3,3),(4,2),(5,1),(6,0) \to \text{ T.N. }=6 \\ x+y=4\Rightarrow (1,3),(2,2),(3,1),(4,0) \to \mathrm{T}.\mathrm{N}=4 \\ x+y=2\Rightarrow (1,1),(2,0) \to \text{ T.N. }=2 \\ x+y=13\Rightarrow (4,9),(5,8),(6,7),(7,6),(8,5),(9,4) \to \mathrm{T}.\mathrm{N}.=6 \\ x+y=11\Rightarrow (2,9),(3,8),(4,7),(5,6),(6,5),(6,5),(7,4),(8,3),(9,2) \to \text{ T.N. }=8 \\ x+y=9\Rightarrow (1,8),(2,7),(3,8),(4,5),(5,4),\dots .(8,1),(9,0) \to \text{ T.N. }=9 \\ x+y=7\Rightarrow (1,8),(2,5),(3,4),\dots (8,1),(7,0) \to \text{ T.N. }=7 \\ x+y=5\Rightarrow (1,4),(2,3),(3,2),(4,1),(5,0) \to \text{ T.N. }=5 \\ \mathrm{x}+\mathrm{y}=20\Rightarrow \text{ Not possible } \\ \mathrm{x}+\mathrm{y}=18\Rightarrow (9,9) \to \mathrm{T}·\mathrm{N}.=1 \\ x+y=16\Rightarrow (7,9),(8,8),(9,7) \to \mathrm{T}.{\mathrm{N}}_{.}=3 \\ x+y=14\Rightarrow (5,9),(6,8),(7,7),(8,6),(9,5) \to \mathrm{T}.\mathrm{N}.=5 \\ x+y=12\Rightarrow (3,9),(4,8),(5,7),(6,6)\dots (9,3) \to \text{ T.N. }=7 \end{gathered}$$