The value of the determinant `|{:(sqrt(6),2i,3+sqrt(6)i),(sqrt(12),sqrt(3)+sqrt(8)i,3sqrt(2)+sqrt(6)i),(sqrt(18),sqrt(2)+sqrt(12)i,sqrt(27)+2i):}|` is (where i=`sqrt(-1)`

Options

A. complex
B. real
C. irrational
D. rational

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} \sqrt{6} & 2i & 3 + \sqrt{6}i \\ \sqrt{12} & \sqrt{3} + \sqrt{8}i & 3\sqrt{2} + \sqrt{6}i \\ \sqrt{18} & \sqrt{2} + \sqrt{12}i & \sqrt{27} + 2i \end{vmatrix} \]

we will follow these steps:

### Step 1: Factor out common terms from the first column
We notice that we can factor out \(\sqrt{6}\) from the first column. This gives us:

\[ D = \sqrt{6} \begin{vmatrix} 1 & 2i & 3 + \sqrt{6}i \\ \sqrt{2} & \sqrt{3} + \sqrt{8}i & 3\sqrt{2} + \sqrt{6}i \\ \sqrt{3} & \sqrt{2} + \sqrt{12}i & \sqrt{27} + 2i \end{vmatrix} \]

### Step 2: Simplify the second and third rows
Next, we simplify the second and third rows by performing row operations. We will perform \(R_2 - \sqrt{2}R_1\) and \(R_3 - \sqrt{3}R_1\):

\[ D = \sqrt{6} \begin{vmatrix} 1 & 2i & 3 + \sqrt{6}i \\ 0 & (\sqrt{3} + \sqrt{8}i - 2i\sqrt{2}) & (3\sqrt{2} + \sqrt{6}i - (3 + \sqrt{6}i)\sqrt{2}) \\ 0 & (\sqrt{2} + \sqrt{12}i - 2i\sqrt{3}) & (\sqrt{27} + 2i - (3 + \sqrt{6}i)\sqrt{3}) \end{vmatrix} \]

### Step 3: Calculate the determinant of the 2x2 matrix
Now we will calculate the determinant of the 2x2 matrix formed by the second and third rows.

Let’s denote the new matrix as:

\[ \begin{vmatrix} 0 & A \\ 0 & B \end{vmatrix} \]

where \(A\) and \(B\) are the entries we calculated in the previous step. The determinant of this matrix is \(0\) since the first column is entirely zeros.

### Step 4: Final calculation
Thus, we have:

\[ D = \sqrt{6} \cdot 0 = 0 \]

### Conclusion
The value of the determinant is:

\[ \boxed{0} \]