jmc

We can write the expression as $$\begin{array}{rl}4x^2-6kxy+(3k^2+2)y^2-4x-4y+6& =x^2-4x+4+2y^2-4y+2+3x^2-6kxy+3k^2{y}^2\\ & =(x^2-4x+4)+2(y^2-2y+1)+3(x^2-2kxy+k^2{y}^2)\\ & =(x-2{)}^2+2(y-1{)}^2+3(x-ky{)}^2.\end{array}$$The only way that this expression can take on the value of 0 is if $x=2,$ $y=1,$ and $x=ky.$ Thus, $k=\boxed{2}.$