Prove that the determinant of a given matrix is proportional to the area of the triangle whose corners are the three points.

Question

For three points in 2D, $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, show that the determinant of

\begin{bmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 2\\ x_3 & y_3 & 3\\ \end{bmatrix}

is proportional to the area of the triangle whose corners are the three points.

I know that a $3\times3$ matrix

\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\\ \end{bmatrix}

has this determinant:

$aei+bfg+cdh-ceg-bdi-afh$

So the above matrix has a determinant of $x_1y_23+y_12x_{3}+1x_2y_3-1y_2x_3-y_1x_23-x_12y_3$

But then I don't know how to prove that this is "proportional to the area of the triangle whose corners are the three points".