The diagonals of a rhombus ABCD intersect at the point M(1,2) and its sides are parallel to the lines $x-\sqrt{3}y+2\sqrt{3}=0and\sqrt{3}x-y+3=0.$ If the vertex A is situated on x-axis, then possible co-ordinates of vertex C are:

Adjacent sides of rhombus ABCD has slope $\sqrt{3}$and $\frac{1}{\sqrt{3}}.$ So angle between them is $\frac{\pi }{3}.$ As diagonal of rhombus is angle bisector of adjacent side, so the equation of diagonal A may be: 

$(y-2)=\pm (x-1)\Rightarrow y=x+1ory=-x+3$ 

$\Rightarrow$Vertex A may be $(-1,0)or(3,0)$

Also length of diagonal AC in both cases is $4\sqrt{2}.$

Now, if $A\equiv (-1,0)$ then vertex C will be $\frac{x+1}{\frac{1}{\sqrt{2}}}=\frac{y-0}{\frac{1}{\sqrt{2}}}=4\sqrt{2}orC\equiv (3,4)$

Also, if $A\equiv (3,0),$ then vertex C will be $\frac{x-3}{\frac{-1}{\sqrt{2}}}=\frac{y-0}{\frac{1}{\sqrt{2}}}=4\sqrt{2}orC\equiv (-1,4)$