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

taking $\sqrt 6$ out from the first column and performing the following operations 
$$R_2\rightarrow R_2 -\sqrt 2 R_1$$
$$R_3 \rightarrow R_3-\sqrt 3 R_1$$
$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$
What should I do next?

The obvious way is to simply find the determinant right away. But the fact that you have asked this must mean that must not be allowed. So to prove this without finding the determinant at all:
Take a factor of $\sqrt3$ from $R_2$ and $\sqrt2$ from $R_3$.
$$\Delta=6\begin{vmatrix} 1 & 2i & 3+\sqrt6 \\
0 & 1 & \sqrt2 i-2\\
0 &1 & \sqrt2 i-3 \end{vmatrix}$$
Now perform $R_2 \to R_2-R_3$
$$=6\begin{vmatrix} 1 & 2i & 3+\sqrt6\\
0 &0 & 1\\
0&1& \sqrt2 i -3 \end{vmatrix}$$
Now $C_2 \to C_2- 2iC_1$,  $C_3 \to C_3 -(3+\sqrt6)C_1$
$$=6\begin{vmatrix} 1 &0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & \sqrt2 i-3 \end{vmatrix}$$
Finally : $C_3 \to C_3 - (\sqrt2 i - 3)C_2 $
$$\Delta = 6\begin{vmatrix} 1 & 0 & 0 \\ 0 & 0& 1 \\ 0&1&0 \end{vmatrix}$$
Now this determinant will obviously be a integral value (as it will be a linear combination of product of integers)
NOTE: I am trying to get rid of the non-integral terms using the 1 0 0 columns. 
