To solve the problem, we need to find the value of the determinant given that \( 2^{a_1}, 2^{a_2}, 2^{a_3}, \ldots, 2^{a_r} \) are in geometric progression (GP).
### Step-by-Step Solution:
1. **Understanding the GP Condition**:
Since \( 2^{a_1}, 2^{a_2}, 2^{a_3} \) are in GP, we know that:
\[
(2^{a_2})^2 = 2^{a_1} \cdot 2^{a_3}
\]
This simplifies to:
\[
2^{2a_2} = 2^{a_1 + a_3}
\]
Since the bases are the same, we can equate the exponents:
\[
2a_2 = a_1 + a_3
\]
2. **Concluding the AP Condition**:
The equation \( 2a_2 = a_1 + a_3 \) indicates that \( a_1, a_2, a_3 \) are in arithmetic progression (AP). This can be generalized for all \( r \) terms, implying that \( a_1, a_2, \ldots, a_r \) are in AP.
3. **Setting Up the Determinant**:
We need to evaluate the determinant:
\[
D = \begin{vmatrix}
a_1 & a_2 & a_3 \\
a_{n+1} & a_{n+2} & a_{n+3} \\
a_{2n+1} & a_{2n+2} & a_{2n+3}
\end{vmatrix}
\]
4. **Using the AP Property**:
Since \( a_1, a_2, \ldots, a_r \) are in AP, we can express:
\[
a_{n+1} = a_1 + (n) d, \quad a_{n+2} = a_1 + (n+1) d, \quad a_{n+3} = a_1 + (n+2) d
\]
\[
a_{2n+1} = a_1 + (2n) d, \quad a_{2n+2} = a_1 + (2n+1) d, \quad a_{2n+3} = a_1 + (2n+2) d
\]
where \( d \) is the common difference.
5. **Row Operations**:
We can perform row operations on the determinant. For instance, we can replace the first row with the sum of the first and third rows:
\[
R_1 \rightarrow R_1 + R_3
\]
This gives us:
\[
D = \begin{vmatrix}
a_1 + a_{2n+1} & a_2 + a_{2n+2} & a_3 + a_{2n+3} \\
a_{n+1} & a_{n+2} & a_{n+3} \\
a_{2n+1} & a_{2n+2} & a_{2n+3}
\end{vmatrix}
\]
6. **Identifying Equal Rows**:
From the property of AP, we can see that:
\[
a_1 + a_{2n+1} = 2a_{n+1}, \quad a_2 + a_{2n+2} = 2a_{n+2}, \quad a_3 + a_{2n+3} = 2a_{n+3}
\]
This means the first row becomes a multiple of the second row.
7. **Conclusion**:
Since two rows of the determinant are now proportional, the determinant evaluates to zero:
\[
D = 0
\]
### Final Answer:
The value of the determinant is \( \boxed{0} \).
