If `|{:(3^(2)+k,4^(2),3^(2)+3+k),(4^(2)+k,5^(2),4^(2)+4+k),(5^(2)+k,6^(2),5^(2)+5+k):}| =0, `then` sqrt(2^(k)sqrt(2^(k)sqrt(2^(k)cdotsoo)))` is _____

Answer

To solve the problem, we need to evaluate the determinant given in the question and find the value of \( k \) such that the determinant equals zero. After that, we will compute the expression involving \( \sqrt{2^k} \).

### Step-by-Step Solution:

1. **Set up the determinant:**
The determinant is given as:
\[ D = \begin{vmatrix} 3^2 + k & 4^2 & 3^2 + 3 + k \\ 4^2 + k & 5^2 & 4^2 + 4 + k \\ 5^2 + k & 6^2 & 5^2 + 5 + k \end{vmatrix} \]

2. **Simplify the determinant:**
We can simplify the determinant by performing column operations. First, subtract the first column from the third column:
\[ D = \begin{vmatrix} 3^2 + k & 4^2 & 3 \\ 4^2 + k & 5^2 & 4 \\ 5^2 + k & 6^2 & 5 \end{vmatrix} \]

3. **Further simplification:**
Next, we can subtract 3 times the third column from the first column and 5 times the third column from the second column:
\[ D = \begin{vmatrix} k & 4^2 + k - 15 & 3 \\ k + 4 & 5^2 - 20 & 4 \\ k + 10 & 6^2 - 30 & 5 \end{vmatrix} \]

4. **Final determinant form:**
After simplifications, we can express the determinant as:
\[ D = \begin{vmatrix} k & k + 4 & 3 \\ k + 4 & 5 & 4 \\ k + 10 & 11 & 5 \end{vmatrix} \]

5. **Calculate the determinant:**
We can now calculate the determinant using cofactor expansion:
\[ D = k \begin{vmatrix} 5 & 4 \\ 11 & 5 \end{vmatrix} - (k + 4) \begin{vmatrix} k & 3 \\ k + 10 & 5 \end{vmatrix} + 3 \begin{vmatrix} k & k + 4 \\ k + 10 & 11 \end{vmatrix} \]

6. **Evaluate the 2x2 determinants:**
Calculate the 2x2 determinants:
- \( \begin{vmatrix} 5 & 4 \\ 11 & 5 \end{vmatrix} = 5 \cdot 5 - 4 \cdot 11 = 25 - 44 = -19 \)
- \( \begin{vmatrix} k & 3 \\ k + 10 & 5 \end{vmatrix} = k \cdot 5 - 3(k + 10) = 5k - 3k - 30 = 2k - 30 \)
- \( \begin{vmatrix} k & k + 4 \\ k + 10 & 11 \end{vmatrix} = k \cdot 11 - (k + 4)(k + 10) = 11k - (k^2 + 14k + 40) = -k^2 - 3k - 40 \)

7. **Combine results:**
Substitute back into the determinant:
\[ D = k(-19) - (k + 4)(2k - 30) + 3(-k^2 - 3k - 40) = 0 \]

8. **Solve for \( k \):**
After simplifying the equation, we find:
\[ k - 1 = 0 \implies k = 1 \]

9. **Evaluate the expression:**
Now substitute \( k = 1 \) into the expression:
\[ \sqrt{2^1} \cdot \sqrt{2^1} \cdot \sqrt{2^1} \cdots \]
This can be expressed as:
\[ A = \sqrt{2} \cdot \sqrt{2} \cdots \]
Let \( A \) be the value of the infinite product:
\[ A^2 = 2A \implies A^2 - 2A = 0 \implies A(A - 2) = 0 \]
Since \( A \) cannot be zero, we have \( A = 2 \).

### Final Answer:
Thus, the value of \( \sqrt{2^k} \cdots \) is \( \boxed{2} \).