= Solution
{c}
Use the <point> $\mathbf p_0=(-2,-2,4)$ and the <orthonormal basis> of directions in the cutting <plane>
$$
\mathbf e_1=\frac{(1,-1,0)}{\sqrt2},\qquad
\mathbf e_2=\frac{(1,1,-1)}{\sqrt3}.
$$
Both <vectors> are <orthogonal> to the <normal vector> $(1,1,2)$, so every <point> of the <plane> has unique <coordinates>
$$
(x,y,z)=\mathbf p_0+s\mathbf e_1+t\mathbf e_2.
$$
Substituting into the <cone> <equation> gives
$$
x^2+y^2-z^2=-8+s^2+\frac{t^2}{3}.
$$
Consequently the intersection has <equation>
$$
\boxed{\frac{t^2}{24}+\frac{s^2}{8}=1.}
$$
Because these are <orthonormal coordinates>, their lengths are the actual lengths in the <plane>. Thus this is an <ellipse> with <semi-major axis> $\sqrt{24}$ and <semi-minor axis> $\sqrt8$, not merely a projected <ellipse>. By <2/a>, its <eccentricity> is
$$
\boxed{e=\sqrt{1-\frac8{24}}=\sqrt{\frac23}.}
$$
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