The correct answer is $\boxed{-3}$.
To find the value of $k$ for which there are infinite number of solutions, we can try to solve the system of equations for $x$ and $y$. However, we can quickly see that this is not possible, since the two equations are not independent. In other words, there is no value of $x$ and $y$ that will satisfy both equations.
This means that the system of equations has no solution, or infinitely many solutions. Intuitively, this makes sense, since the two equations are essentially saying the same thing: $3x+2ky=-2$ and $kx+6y=2$. So, no matter what value we choose for $x$, we can always find a value for $y$ that will satisfy both equations.
More formally, we can say that the system of equations has infinitely many solutions if the determinant of the coefficient matrix is equal to zero. In this case, the coefficient matrix is $$\begin{bmatrix} 3 & 2k \\ k & 6 \end{bmatrix}$$ and the determinant is $$3\times 6 – 2k\times k = 18 – 2k^2$$
So, the system of equations has infinitely many solutions if $18 – 2k^2 = 0$. This means that $k^2 = 9$, or $k = \pm 3$.
However, only the value $k=-3$ makes sense in the context of the original system of equations. This is because the original system of equations is asking us to find values for $x$ and $y$ that satisfy both equations. If we choose $k=3$, then the first equation becomes $3x+6y=-2$, which has no solution. So, the only value of $k$ for which there are infinite number of solutions is $\boxed{-3}$.