1. Let us find the probability of drawing an odd or even number from each box.

• Box 1 has cards $\{1,2,3\}$

  • Odd numbers: $1,3$ $\Rightarrow 2$ cards
  • Even numbers: $2$ $\Rightarrow 1$ card

  Hence, $P(O_1)=\frac{2}{3}, \qquad P(E_1)=\frac{1}{3}$

• Box 2 has cards $\{1,2,3,4,5\}$

  • Odd numbers: $1,3,5$ $\Rightarrow 3$ cards
  • Even numbers: $2,4$ $\Rightarrow 2$ cards

  Hence, $P(O_2)=\frac{3}{5}, \qquad P(E_2)=\frac{2}{5}$

• Box 3 has cards $\{1,2,3,4,5,6,7\}$

  • Odd numbers: $1,3,5,7$ $\Rightarrow 4$ cards
  • Even numbers: $2,6$? Actually even numbers are $2,4,6$ $\Rightarrow 3$ cards

  Hence, $P(O_3)=\frac{4}{7}, \qquad P(E_3)=\frac{3}{7}$

2. The sum $x_1+x_2+x_3$ is odd when the number of odd terms is odd.

For three numbers, this happens in two cases:

• exactly one odd and two even,
• all three odd.

So, $P(\text{odd sum}) = P(OEE)+P(EOE)+P(EEO)+P(OOO)$

3. Compute each term:

• $P(OEE)=P(O_1)P(E_2)P(E_3)=\frac{2}{3}\cdot\frac{2}{5}\cdot\frac{3}{7}=\frac{4}{35}$

• $P(EOE)=P(E_1)P(O_2)P(E_3)=\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{3}{7}=\frac{3}{35}$

• $P(EEO)=P(E_1)P(E_2)P(O_3)=\frac{1}{3}\cdot\frac{2}{5}\cdot\frac{4}{7}=\frac{8}{105}$

• $P(OOO)=P(O_1)P(O_2)P(O_3)=\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{4}{7}=\frac{8}{35}$

4. Add them:

$P(\text{odd sum})=\frac{4}{35}+\frac{3}{35}+\frac{8}{105}+\frac{8}{35}$

Convert to denominator $105$:

$\frac{4}{35}=\frac{12}{105}, \quad \frac{3}{35}=\frac{9}{105}, \quad \frac{8}{35}=\frac{24}{105}$

Thus, $P(\text{odd sum})=\frac{12}{105}+\frac{9}{105}+\frac{8}{105}+\frac{24}{105}=\frac{53}{105}$

5. Therefore, the correct option is $\boxed{\frac{53}{105}}$ which is Option B.