Try to put these equations in more familiar forms to know for sure what shape they represent. For example, 3) could be written as $y + z = 2$. In general, some algebraic manipulation takes these polynomial equations into the canonical expressions that help us gain some intuition and make a mental image of the shapes. Hint for 1) and 2): complete the squares.
1) is an elliptic cylinder, and 2) is a parabolic cylinder and 3) is a plane.
Added for 1): From $x^2 + 2y^2 - 6x + 4y + 7 = 0$ we see that we have quadratic terms in $x$ and $y$. I would try to isolate $x$ and $y$ like this: $(x^2 - 6x) + 2(y^2 + 2y) + 7 = 0$. Now, we complete the squares: $(x^2 - 6x + 9) - 9 + 2(y^2 + 2y + 1) - 2 + 7 = 0$ (we add and substract the necessary quantities). Then $(x-3)^2 + 2(y+1)^2 = 4$. Dividing through by $4$ we have $\frac{(x-3)^2}{4} + \frac{(y+1)^2}{2} = 1$, which is the equation of an ellipse in the $xy$-plane. Since $z$ is absent in the equation, for all $z$ the shape will be the same, and we have an elliptic cylinder with axis parallel to the $z$ axis, i.e., in a "vertical" position.
Added: $(x-1)^2 + (y-3)^2 + 6z = -10$ isn't a cylinder. Note that since all variables are involved in the equation, you could rearrange to get: $z = -\frac{1}{6}(x-1)^2 - \frac{1}{6}(y-3)^2 -\frac{10}{6}$, which is the graph of an "upside down" paraboloid.