Parameterization of a triangle in $\mathbb R^3$ (edge and surface)

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Let $S$ be a triangle with vertices at A=$(1,2,0)$, B=$(2,3,2)$, and C=$(0,0,4)$. Find a parametrization of the triangle and its contour.

I have set up the parametric equations for each edge

$\begin{pmatrix} 2 \\ 3 \\ 2\end{pmatrix}$-$\begin{pmatrix} 1 \\ 2 \\ 0\end{pmatrix}$=$\begin{pmatrix} 1 \\ 1 \\ 2\end{pmatrix}$

first edge: $r_1(t)=\begin{pmatrix} 1 \\ 2 \\ 0\end{pmatrix} + t\cdot \begin{pmatrix} 1 \\ 1 \\ 2\end{pmatrix}=\begin{pmatrix} 1+t \\ 2+t \\ 0+2t\end{pmatrix}$

likewise for the others

$r_2(t)=\begin{pmatrix} 2-2t \\ 3-3t \\ 2+2t\end{pmatrix}$

$r_3(t)=\begin{pmatrix} 0+t \\ 0+2t \\ 0-4t\end{pmatrix}$

for each $t \in [0,1]$

What can I do now? How to put them together, and how to get the parametrization of the whole surface and not only the edges?

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What you have done is all you need to do for the edges. You could parametrize so you could go from $0 \leq t \leq 1, 2 \leq t \leq 2, 2 \leq t \leq 3$.

$r_1(t)= (1, 2, 0) + (1, 1, 2)t = (1 + t, 2 + t, 2t) \,$ for $( 0 \leq t \leq 1)$

$r_2(t)= (2, 3, 2) + (-2, -3, 2) (t-1) = (4-2t, 6-3t, 2t) \,$ for $( 1 \leq t \leq 2)$

$r_3(t)= (0, 0, 4) + (1, 2, -4) (t-2) = (-2 + t, -4 + 2t, 12-4t) \,$ for $( 2 \leq t \leq 3)$

Now if you have to parametrize the surface, find the equation of the plane it is on. First take two directional vectors you found above and do a cross product to find the normal vector to the plane-

$(-2, -3, 2) \times (1, 1, 2) = (-8, 6, 1)$

Take any point on the surface say $(0, 0, 4)$ then we have,

$ -8(x-0) + 6(y-0) + (z-4) = 0$

Parametrization of the surface of the plane the given triangle is on,

$ \,z = 4 + 8x - 6y$

EDIT:

On the parametrization that you asked,

If you take parametrization of two of your edges, $r_1(s) = (1 + s, 2 + s, 2s) \,(0 \leq s \leq 1)$

$r_2(t) = (2 - 2t, 3 - 3t, 2 + 2t) \,(0 \leq t \leq 1)$

You can combine these two to parametrize your triangle.

$r(s, t) = (1 + s - 2t, 2 + s - 3t, 2s + 2t) \, \, (0 \leq s, t \leq 1)$

If you take cross product $r_t \times r_s$, you will get the normal vector $(-8, 6, 1)$ which is what I used for other parametrization.