Prove $\vec{r}=c\vec{u}+d\vec{v}+e\vec{w}$ with $c,d,e\in[0,1]$ and $c+d+e\leq1$ represent the tetrahedron formed by the vectors $\vec{u},\vec{v}$ and $\vec{w}$
Attempt 1
Slicing the tetrahedron into triangles, say $\Delta PQT$ and shift origin to say $R(t\vec{w})$.
For $\vec{r}$ to be any point on the triangle $\Delta{PQT}$, ie., for $\vec{r}-t\vec{w}$ to lie in the $\Delta PQT$, where $p,q,t\in[0,1]$
$\vec{r}-t\vec{w}=\lambda(q\vec{v}-t\vec{w})+\mu(p\vec{u}-t\vec{w})$ with $0\leq\lambda,\mu\leq 1$ and $\lambda+\mu\leq 1$
$\implies \vec{r}=\mu p\vec{u}+\lambda q\vec{v}+t(1-\lambda-\mu)\vec{w}=c\vec{u}+d\vec{v}+e\vec{w}$
$$ p\mu=c,\quad q\lambda=d,\quad t-t(\lambda+\mu)=e\\ c+d+e=p\mu+q\lambda+t[1-(\lambda+\mu)]=\mu(p-t)+\lambda(q-t)+t $$
Why am I not getting the required conditions in my attempt ?
Note: $c\vec{u}+d\vec{v}+e\vec{w}$ represents $\Delta ABC$ with the conditions $c,d,e\in[0,1]$ and $c+d+e=1$, which can be easily proved by a similar attempt.
Attempt 2
For $\vec{r}$ to be any point on the triangle $\Delta{ABC}$, ie., for $\vec{r}-\vec{w}$ to lie in the $\Delta ABC$.
$\vec{r}-\vec{w}=\lambda(\vec{v}-\vec{w})+\mu(\vec{u}-\vec{w})$ with $0\leq\lambda,\mu\leq 1$ and $\lambda+\mu\leq 1$
$\implies \vec{r}=\mu \vec{u}+\lambda \vec{v}+t(1-\lambda-\mu)\vec{w}=c\vec{u}+d\vec{v}+e\vec{w}$
$$ \mu=c,\quad \lambda=d,\quad 1-(\lambda+\mu)=e\\ c+d+e=\mu+\lambda+t[1-(\lambda+\mu)]=1 \text{ and } c,d,e\in[0,1] $$
For any point inside the tetrahedron OABC, the position vector $\color{red}{\vec{p}=\alpha\vec{r}}=\alpha c\vec{u}+\alpha d\vec{v}+\alpha e\vec{w}=\psi\vec{u}+\phi\vec{v}+\eta\vec{w}$ where $\boxed{\psi+\phi+\eta=\alpha(c+d+e)\leq 1\text{ and }\psi,\phi,\eta\in[0,1]}$
Can this be considered as a decent proof which shows the statement ?