Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either left or down, both equally likely. Which of the following is closest to the probability that the objects meet?

$ \textbf{(A)}\ 0.10 \qquad
\textbf{(B)}\ 0.15 \qquad
\textbf{(C)}\ 0.20 \qquad
\textbf{(D)}\ 0.25 \qquad
\textbf{(E)}\ 0.30$
1. **Determine the possible meeting points:**
   - Object \( A \) starts at \((0,0)\) and moves either right or up.
   - Object \( B \) starts at \((5,7)\) and moves either left or down.
   - For the objects to meet, they must end up at the same coordinates \((x, y)\).

2. **Calculate the number of steps each object takes:**
   - Object \( A \) must take 5 steps to the right and 7 steps up to meet at any point \((x, y)\).
   - Object \( B \) must take 5 steps to the left and 7 steps down to meet at any point \((x, y)\).

3. **Identify the possible meeting points:**
   - Since both objects take 6 steps, the possible meeting points are where the sum of the steps taken by each object equals 6.
   - The possible meeting points are:
     \[
     (5,1), (4,2), (3,3), (2,4), (1,5), (0,6)
     \]

4. **Calculate the number of ways to reach each meeting point:**
   - For each meeting point \((x, y)\), calculate the number of ways object \( A \) and object \( B \) can reach that point.
   - The number of ways object \( A \) can reach \((x, y)\) is given by \(\binom{6}{x}\).
   - The number of ways object \( B \) can reach \((x, y)\) is given by \(\binom{6}{7-y}\).

5. **Sum the number of ways for all meeting points:**
   \[
   \sum_{i=0}^{5} \binom{6}{i} \binom{6}{5-i} = \binom{6}{0}\binom{6}{5} + \binom{6}{1}\binom{6}{4} + \binom{6}{2}\binom{6}{3} + \binom{6}{3}\binom{6}{2} + \binom{6}{4}\binom{6}{1} + \binom{6}{5}\binom{6}{0}
   \]

6. **Calculate the total number of ways:**
   \[
   \binom{6}{0}\binom{6}{5} + \binom{6}{1}\binom{6}{4} + \binom{6}{2}\binom{6}{3} + \binom{6}{3}\binom{6}{2} + \binom{6}{4}\binom{6}{1} + \binom{6}{5}\binom{6}{0} = 1 \cdot 6 + 6 \cdot 15 + 15 \cdot 20 + 20 \cdot 15 + 15 \cdot 6 + 6 \cdot 1 = 792
   \]

7. **Calculate the total number of possible paths:**
   - Each object has \(2^6\) possible paths.
   - The total number of possible paths for both objects is \((2^6)^2 = 2^{12} = 4096\).

8. **Calculate the probability:**
   \[
   \text{Probability} = \frac{792}{4096} \approx 0.193359375 \approx 0.20
   \]

The final answer is \(\boxed{0.20}\)