Let both the series $$a_1, a_2, a_3, \dots$$ and $$b_1, b_2, b_3, \dots$$ be in arithmetic progression such that the common differences of both the series are prime numbers. If $$a_5 = b_9$$, $$a_{19} = b_{19}$$, and $$b_2 = 0$$, then $$a_{11}$$ equals

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1. **Define the Series:**  
   Let the common differences of the series $$a_1, a_2, a_3, \dots$$ and $$b_1, b_2, b_3, \dots$$ be $$p$$ and $$q$$ respectively (both are prime numbers). The terms of the series can be expressed as:  
   - $$a_n = a_1 + (n-1)p$$  
   - $$b_n = b_1 + (n-1)q$$

2. **Given Conditions:**  
   - $$a_5 = b_9$$:  
     $$
     a_1 + 4p = b_1 + 8q.
     $$  
   - $$a_{19} = b_{19}$$:  
     $$
     a_1 + 18p = b_1 + 18q.
     $$  
   - Subtracting these equations:  
     $$
     (a_{19} - a_{5}) = (b_{19} - b_{9}) \implies 14p = 10q.
     $$  
   - Simplify:  
     $$
     \frac{p}{q} = \frac{10}{14} = \frac{5}{7}.
     $$

3. **Prime Values of $$p$$ and $$q$$:**  
   Since both $$p$$ and $$q$$ are prime numbers:  
   - $$p = 5$$ and $$q = 7$$.

4. **Calculate Terms Using Given Condition ($$b_2 = 0$$):**  
   From the formula for terms in the second sequence:  
   $$
   b_n = b_1 + (n-1)q.
   $$  
   Substituting $$b_2 = 0$$:  
   $$
   b_1 + q = 0 \implies b_1 = -7.
   $$

5. **Find $$a_{11}$$:**  
   Using the formula for terms in the first sequence:  
   $$
   a_{11} = a_5 + 6p.
   $$  
   Substituting values:  
   - From $$a_5 = b_9$$:  
     $$
     a_5 = b_9 = b_2 + 7q \implies a_5 = 0 + 7 \cdot 7 = 49.
     $$  
   - Substituting into $$a_{11}$$:  
     $$
     a_{11} = 49 + 6 \cdot 5 = 49 + 30 = 79.
     $$

Thus, the value of $$a_{11}$$ is **79**.

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