= Solution
{c}
Let $D$ be the filled projected <ellipse>. The part of the <plane> bounded by $C$ is the <parametrized surface>
$$
\mathbf R(x,y)=(x,y,1-x-y),\qquad (x,y)\in D.
$$
Its <tangent vectors> satisfy
$$
\mathbf R_x=(1,0,-1),\qquad \mathbf R_y=(0,1,-1),\qquad
\|\mathbf R_x\times\mathbf R_y\|=\|(1,1,1)\|=\sqrt3.
$$
The <surface area> is therefore
$$
\boxed{\mathcal A(C)=\iint_D\sqrt3\,dx\,dy=\sqrt3\frac{8\pi}{3\sqrt3}=\frac{8\pi}{3}.}
$$
This also illustrates the <orthogonal projection area formula>: the <unit normal> to the <plane> has vertical component $1/\sqrt3$, so projection multiplies <area> by $1/\sqrt3$.
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