Answer
Vector parametrization:
${\bf{r}}\left( \theta \right) = \left( {1 - 3\cos \theta - 8\sin \theta ,3\cos \theta ,8\sin \theta } \right)$, ${\ \ \ }$ for $0 \le \theta \le 2\pi $
Work Step by Step
Consider the elliptical cylinder ${\left( {\frac{y}{3}} \right)^2} + {\left( {\frac{z}{8}} \right)^2} = 1$. It can be parametrized by
$\left( {y,z} \right) = \left( {3\cos \theta ,8\sin \theta } \right)$, ${\ \ }$ for $0 \le \theta \le 2\pi $
Notice that it satisfied the equation ${\left( {\frac{y}{3}} \right)^2} + {\left( {\frac{z}{8}} \right)^2} = 1$.
Substituting $y = 3\cos \theta $ and $z = 8\sin \theta $ in $x + y + z = 1$ gives
$x + 3\cos \theta + 8\sin \theta = 1$
$x = 1 - 3\cos \theta - 8\sin \theta $
Thus, the parametrization of the intersection of the plane $x + y + z = 1$ and the elliptical cylinder ${\left( {\frac{y}{3}} \right)^2} + {\left( {\frac{z}{8}} \right)^2} = 1$ is given by
${\bf{r}}\left( \theta \right) = \left( {1 - 3\cos \theta - 8\sin \theta ,3\cos \theta ,8\sin \theta } \right)$, ${\ \ \ }$ for $0 \le \theta \le 2\pi $
