Any convex combination of positive numbers must be greather than 0, hence
{e_1,e_2, e_3} must be in the hull.
Suppose that some of the last 3 vector doesn't exist in the hull, for example $(1/2, 1/2, 1)$.
Let $\Delta := conv({e_1,e_2, e_3})$. It is known that vector $(1/2, 1/2, 1)$ can be written as convex combination of vector $v \in \Delta$ and vectors: $(1/2, 1, 1/2)$ , $(1, 1/2, 1/2)$. So there exists
$\alpha_1,\alpha_2,\alpha_3 \geq 0$, $\alpha_1 + \alpha_2 +\alpha_3 =1$,
$ v=(\alpha_1,\alpha_2,\alpha_3)$,
$\beta_1, \beta_2, \beta_3 \geq 0$, $\beta_1 + \beta_2 + \beta_3 = 1$ such that
$\beta_1 (\alpha_1,\alpha_2,\alpha_3) + \beta_2 (1/2, 1, 1/2) + \beta_3 (1, 1/2, 1/2) = (1/2, 1/2, 1)$.
We can write 3 equations:
$$
\begin{cases}
\beta_1 \alpha_1 +\beta_2 1/2 + \beta_3 = 1/2 \\
\beta_1 \alpha_2 +\beta_2 + \beta_3 1/2 = 1/2 \\
\beta_1 \alpha_3 +\beta_2 1/2 + \beta_3 1/2 = 1
\end{cases}
$$
If $\beta_1 > 0$,
after summing these 3 equations by sides, we get:
$\beta_1 (\alpha_1 + \alpha_2 +\alpha_3)+2(\beta_1+\beta_2) = 2 $
but $\beta_1 (\alpha_1 + \alpha_2 +\alpha_3)+2(\beta_1+\beta_2) =
\beta_1 +2(\beta_1+\beta_2) < 2 \beta_1 +2(\beta_1+\beta_2) = 2$
- contraditon.
If $\beta_1 = 0$ we have:
$$
\begin{cases}
\beta_2 1/2 + \beta_3 = 1/2 \\
\beta_2 + \beta_3 1/2 = 1/2 \\
\beta_2 1/2 + \beta_3 1/2 = 1
\end{cases}
$$
after adding 2 first equations by sides, we get:
$$
\begin{cases}
\beta_2 3/2 + \beta_3 3/2 = 1 \\
\beta_2 1/2 + \beta_3 1/2 = 1
\end{cases}
$$
and these equations are contraty.