Find the smallest positive integer $n$ such that the number $2^n + 2^8 + 2^{11}$ is a perfect square.

(A): $8$,  (B): $9$,  (C): $11$,  (D): $12$, (E) None of the above.
1. We start with the expression \(2^n + 2^8 + 2^{11}\) and need to find the smallest positive integer \(n\) such that this expression is a perfect square.
2. Let's rewrite the expression in a more convenient form:
   \[
   2^n + 2^8 + 2^{11}
   \]
   Notice that \(2^8 = 256\) and \(2^{11} = 2048\). We can factor out the smallest power of 2 from the terms:
   \[
   2^n + 2^8 + 2^{11} = 2^8(2^{n-8} + 1 + 2^3)
   \]
   Simplifying further:
   \[
   2^8(2^{n-8} + 1 + 8)
   \]
   We need \(2^{n-8} + 9\) to be a perfect square. Let \(k^2 = 2^{n-8} + 9\), where \(k\) is an integer.
3. Rearrange the equation to solve for \(2^{n-8}\):
   \[
   2^{n-8} = k^2 - 9
   \]
   Factor the right-hand side:
   \[
   2^{n-8} = (k-3)(k+3)
   \]
   Since \(2^{n-8}\) is a power of 2, both \((k-3)\) and \((k+3)\) must be powers of 2. Let \(k-3 = 2^a\) and \(k+3 = 2^b\), where \(a < b\). Then:
   \[
   2^b - 2^a = 6
   \]
4. We need to find powers of 2 that satisfy this equation. Let's check possible values for \(a\) and \(b\):
   - If \(a = 1\) and \(b = 3\):
     \[
     2^3 - 2^1 = 8 - 2 = 6
     \]
     This works. So, \(a = 1\) and \(b = 3\).
5. Now, we have:
   \[
   k - 3 = 2^1 = 2 \quad \text{and} \quad k + 3 = 2^3 = 8
   \]
   Solving for \(k\):
   \[
   k = 2 + 3 = 5
   \]
6. Substitute \(k = 5\) back into the equation \(2^{n-8} = k^2 - 9\):
   \[
   2^{n-8} = 5^2 - 9 = 25 - 9 = 16 = 2^4
   \]
   Therefore:
   \[
   n - 8 = 4 \implies n = 12
   \]

The final answer is \(\boxed{12}\)