Prove that the value of $\Delta$ is an integer for the given determinant $\Delta=\begin{vmatrix} \sqrt 6&amp; 2i&amp; 3+\sqrt 6 \\ \sqrt{12}&amp;\sqrt 3 +\sqrt8i &amp;3\sqrt 2 +\sqrt 6i \\ \sqrt{18} &amp;\sqrt 2+ \sqrt {12}i &amp;\sqrt {27}+2i \end{vmatrix}$


taking $\sqrt 6$ out from the first column and performing the following operations 

$$R_2\rightarrow R_2 -\sqrt 2 R_1$$

$$R_3 \rightarrow R_3-\sqrt 3 R_1$$

$$\sqrt 6\begin {vmatrix} 1&amp;2i&amp;3+\sqrt 6 \\\ 0&amp;\sqrt 3&amp;\sqrt 6i-2\sqrt 3 \\\ 0&amp;\sqrt 2&amp;2i-3\sqrt 2 \end {vmatrix}$$

What should I do next?
From here

$$\sqrt 6\begin {vmatrix} 1&amp;2i&amp;3+\sqrt 6 \\\ 0&amp;\sqrt 3&amp;\sqrt 6i-2\sqrt 3 \\\ 0&amp;\sqrt 2&amp;2i-3\sqrt 2 \end {vmatrix}$$

Pull out a factor of $\sqrt{3}$ from the second row and a factor of $\sqrt{2}$ from the third to get

$$6\begin {vmatrix} 1&amp;2i&amp;3+\sqrt 6 \\\ 0&amp; 1 &amp;\sqrt 2i-2 \\\ 0&amp; 1 &amp; \sqrt{2}i-3 \end {vmatrix}$$

Now you can compute it easily by expanding along the first row or by subtracting the second row from the third. Lets subtract.

$$6\begin {vmatrix} 1&amp;2i&amp;3+\sqrt 6 \\\ 0&amp; 1 &amp;\sqrt 2i-2 \\\ 0&amp; 1 &amp; \sqrt{2}i-3 \end {vmatrix} = 6\begin {vmatrix} 1&amp;2i&amp;3+\sqrt 6 \\\ 0&amp; 1 &amp;\sqrt 2i-2 \\\ 0&amp; 0 &amp; -1 \end {vmatrix}$$

The determinant is just the product of the diagonal entries, so $6 \cdot 1 \cdot 1 \cdot -1 = -6$.
Expand along the first column. Thanks to the two zeroes there, you will only have $\sqrt{6}\cdot1\cdot(2\times2\text{ determinant})$. And you can calculate that small determinate directly.