$$\left|\begin{matrix} x & y & z & 1 \\ x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \end{matrix}\right|=0$$
This is the equation of the plane that I have (in 3D). Now I need to find the area of projection of this plane on the $xy,yz$ and $zx$ planes (in the direction of positive axes only). How should I find it ?
Please give your answers in determinant form.
Please note: I just started learning 3d geometry today. So don't mind if my question is too silly :-P...
If you have a polygon in space and want to find its projection to a particular plane, what you do is project each of the vertices to that plane and draw the corresponding polygon there. You can then compute the area of the projection by working in the plane directly.
In the case of, for example, the $xy$-plane, the initial projection into that plane consists simply of throwing away the $z$ coordinates (you when you vary the $z$ coordinate, the point moves along a line parallel to the $z$-axis, and the $x$ and $y$ coordinates you already have describe the intersection of that line with the $xy$-plane).
The determinant $$\frac12\left|\begin{matrix}x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ x_4 & y_4 & 1 \end{matrix}\right|$$ reduces to the shoelace formula for the area of a plane triangle when you expand it by minors along the last column. (Alternatively, it is probably possible with a bit of algebraic manipulation to show that it is equal to a determinant that represents the volume of a 3D parallelepiped where the original triangle is half of its basis and the height is $1$, but that doesn't sound very enlightening to me).
The original 4-by-4 determinant in the question is a systematic way to find an equation for the infinite plane that contains the three points $(x_i,y_i,z_i)$. An equation for that plane is, however, not directly useful if what you're interested in is the area of particular bounded subsets of it.