= Solution
{c}
Eliminate $z$ using the <plane> <equation>. The <orthogonal projection> consists exactly of the <points> satisfying
$$
x^2+xy+y^2+x+y=1,
$$
since each such $(x,y)$ lifts uniquely to $(x,y,1-x-y)$ on the intersection. Translating to the <center of an ellipse> with $X=x+1/3$ and $Y=y+1/3$ gives
$$
X^2+XY+Y^2=\frac43.
$$
The <symmetric matrix> of this <quadratic form> is $M=\left(\begin{smallmatrix}1&1/2\\1/2&1\end{smallmatrix}\right)$, whose <eigenvalues> are $3/2$ and $1/2$. Thus it is a <symmetric positive-definite matrix>, and the projected curve is an <ellipse>. The <quadratic-form ellipse area>, with the right-hand side scaled by $4/3$, is
$$
\boxed{\mathcal A(C')=\frac{(4/3)\pi}{\sqrt{3/4}}=\frac{8\pi}{3\sqrt3}.}
$$
For example, the <orthogonal change of coordinates> $u=(X+Y)/\sqrt2$, $v=(X-Y)/\sqrt2$ gives semiaxes $2\sqrt2/3$ and $2\sqrt6/3$, with the same product.
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