Is there a way to cut a an ellipsoid with a plane such that it gives a circle? I'm trying to answer this

In $\Bbb {R^3} $ consider the ellipsoid:
  $2x^2+3y^2+4z^2=1$ It exists a subspace of dimension 2 which intersection with the ellipsoid is a circle. Justify any answer.

I know that I must search for a plane since these are the only linear subspaces of $\Bbb {R^3} $ of dimension 2, but I don't know what else to do to check if such a plane exists. any hint would be much appreciated. Thanks!
 A: Let $a,b,c$ be any 3 positive numbers such that $0 < a < b < c$. 
Let $\mathcal{E}$ and $\mathcal{S}$ be the ellipsoid and sphere defined by
$$\mathcal{E} = \{ (x,y,z) : ax^2+by^2 + cz^2 = 1 \}
\quad\text{ and }\quad
\mathcal{S} = \{ (x,y,z) : b(x^2+y^2+z^2) = 1 \}$$
For any point $(x,y,z) \in \mathcal{E} \cap \mathcal{S}$, we have
$$ax^2 + by^2 + cz^2 = 1 = b(x^2+y^2+z^2)
\implies (c-b)z^2 = (b-a)x^2
\iff z = \pm k x
$$
where $k = \sqrt{\frac{b-a}{c-b}}$. The RHS is the equation for a pair of planes
$P_{\pm} = \{ (x,y,z) : z = \pm k x \}$.
This implies
$$\mathcal{S} \cap \mathcal{E} \subset \mathcal{P}_{+} \cup \mathcal{P}_{-}$$
Converse, for any point $(x,y,z) \in \mathcal{S} \cap (\mathcal{P}_{+} \cup \mathcal{P}_{-})$, a similar argument tell us $ax^2 + by^2 + cz^2 = 1$.
This means we also has
$$\mathcal{S} \cap (\mathcal{P}_{+} \cup \mathcal{P}_{-}) \subset \mathcal{E}$$
Combine these two piece of information, we get
$$\mathcal{S} \cap \mathcal{E}
= \mathcal{S} \cap ( \mathcal{P}_{+} \cup \mathcal{P}_{-} )
= ( \mathcal{S} \cap \mathcal{P}_{+} ) \cup (\mathcal{S} \cap \mathcal{P}_{-})$$
Since the intersection of a sphere and a plane is a circle, we find 
$\mathcal{S} \cap \mathcal{E}$ is the union of two circles 
$\mathcal{S} \cap \mathcal{P}_{+}$ and $\mathcal{S} \cap \mathcal{P}_{-}$.
We can repeat essentially the same argument to other combinations of $\mathcal{S}$, $\mathcal{E}$ and $\mathcal{P}_{\pm}$. At the end, we have something like
$$\mathcal{E} \cap \mathcal{P}_{+} = \mathcal{S} \cap \mathcal{P}_{+}
\quad\text{ and }\quad
\mathcal{E} \cap \mathcal{P}_{-} = \mathcal{S} \cap \mathcal{P}_{-}$$
Substitute $(a,b,c)$ by $(2,3,4)$, we find the intersection of
the ellipsoid $$\{ (x,y,z) : 2x^2 + 3y^2 + 4z^2 = 1 \}$$ with the two planes $z = \pm x$ are two circles of radius $\frac{1}{\sqrt{b}} = \frac{1}{\sqrt{3}}$.
A: The problem becomes easier if you go to parametric equations:
$$x=a\cos(u)\cos(v)\\
y=b\cos(u)\sin(v)\\
z=c\sin(v)$$
you know that the equation of a circle is $x^2+y^2+z^2=r^2$, being $r$ the radius of the circle. You only have to find $v$ for example, such that the equation of the circle holds:
$$r^2=(a^2\cos(v)^2+b^2\sin(v)^2)\cos(u)^2+c^2\sin(u)^2$$
If you want to obtain a constant independent of $u,v$, you have that  $(a^2\cos(v)^2+b^2\sin(v)^2)=c^2$. If you consider that $c<b,c$ (if this is not true, you can change the parametrization of the ellipsoid, and follow a similar reasoning), there will be always a solution to the equation and the radius of the circle will be $r=c$. Now, if you have the $v$, you can return back to your ellipsoid parametrized and rebuild the plane which leads to such circle (a plane is defined by three points).
This is only one of the infinity solutions (see parametrization chapter of https://en.wikipedia.org/wiki/Ellipsoid). You can obtain the rest of solution by taking parallel planes to the one you have computed.
