$x+2y=6$ is rearranged to be $y=-\frac{1}{2}x+3$.

Since the slope is negative, we eliminate Choice J.

And it intersects the y axis on (0, 3).

Therefore, Choice F is eliminated.

Plug in (0, 0) into $3x^2+3y^2>12$ and the inequality fails, therefore the shaded area must be outside of the circle. Choice G is eliminated.

Plug in (0, 0) into $x+2y \leq 6$ and the inequality succeeds, therefore the shaded area must cover (0, 0). Choice K is correct.