I have learned the formula of area of triangle knowing the vertices: $$S=\frac{1}{2}\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}$$ But I wonder why is a determinant of a matrix used to calculate the area.
So I have 3 questions:
Informal intuition:
The implicit equation of a straight line is
$$ax+by+c=0.$$
If you are given three points and want to express that they are aligned, you can write the system of three equations in the three unknowns $a,b,c$ and express that it has a non-trivial solution. This requires that the following condition be fulfilled:
$$\Delta:=\begin{vmatrix}x_0&y_0&1\\x_1&y_1&1\\x_2&y_2&1\end{vmatrix}=0.$$
So this $\Delta$ is in a way a measure of the misalignment of the points. In addition, it is a polynomial of the second degree in $x_i,y_i$ and is a quantity with the dimension of an area. If you dilate one of the coordinates by a constant factor, the $\Delta$ grows linearly. So it is a good candidate to be proportional to the area of the triangle. Notice that this area is signed, depending on the orientation of the triangle.
The same reasoning holds in the 3D case.
I think of this formula as a pretty result that is derived by algebraic manipulations, starting from more primitive and more basic formulas, using relations between determinants and geometry.
As a first step, let's massage the formula a little bit. By doing two row operations, namely subtracting the first row from rows $2$ and $3$, the formula is converted into $$\frac{1}{2} \begin{vmatrix}x_0&y_0&1\\(x_1-x_0)&(y_1-y_0)&0\\(x_2-x_0)&(y_2-y_0)&0\end{vmatrix} = \frac{1}{2} \begin{vmatrix}(x_1-x_0)&(y_1-y_0)\\(x_2-x_0)&(y_2-y_0)\end{vmatrix} $$ Incidently, regarding your third bullet point, perhaps you can see from these algebraic manipulations so far how the final column of $1$'s was used.
Let
In the equation above, the left hand side is the formula for the points $P_0,P_1,P_2$ and the right hand side is the formula for $O,Q_1,Q_2$. So we have proved that those two formulas are equal.
Also, the translation of the plane that moves $P_0$ to $O$ is a rigid motion taking the $P_0,P_1,P_2$ triangle to the $O,Q_1,Q_2$ triangle. Since rigid motions preserve area, we have proved that those two triangles have equal area.
Therefore, we have reduced the problem to the following: given a triangle with one vertex at $O$ and two other vertices at $Q_1 = (a_1,b_1)$ and $Q_2=(a_2,b_2)$, its area is given by the formula $$\frac{1}{2} \begin{vmatrix} a_1 & b_1 \\ a_1 & b_2 \end{vmatrix} $$
Now we can reduce this one further step, namely to a parallelogram problem: if you rotate this triangle $180^\circ$ around the midpoint of the $Q_1,Q_2$ edge you get another triangle which, when taken together with the original triangle, is a parallelogram. We say that this parallelogram is spanned by the line segments $\overline{O Q_1}$ and $\overline{O Q_2}$. The two triangles whose union forms this parallelogram have the same area, because rotation is a rigid motion. Also, the two triangles have no overlap of their areas. The parallelogram therefore has area exactly two times the area of the $O Q_1 Q_2$ triangle.
So now the problem is reduced one further step: prove that the area of the parallelogram spanned by $\overline{O Q_1}$ and $\overline{O Q_2}$ is $$\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} $$ This is the "more basic formula" that I mentioned in my opening sentence.
But this formula still has to be proved, and here again one can derive it from a still more basic formula by making use of determinants and geometry.
Let's do another rigid motion of the plane, this time a rotation around the origin which rotates the ray through $(a_1,b_1)$ to the positive $x$-axis. Under this rotation, the point $(a_1,b_1)$ goes to some point $S_1 = (c_1,0)$ on the positive $x$-axis, and the point $(a_2,b_2)$ goes to some point $S_2 = (d_1,d_2)$. This rotation takes the parallelogram spanned by $\overline{O Q_1}$ and $\overline{O Q_2}$ to the parallelogram spanned by $\overline{O S_1}$ and $\overline{O S_2}$, and so the areas of these two parallelograms are equal.
Also, the formulas for those two parallelograms are equal, which can be seen as follows. Letting $\theta$ be the angle between the two rays mentioned above, this rotation is represented by the matrix $R = \begin{pmatrix} \cos\theta & \sin \theta \\ -\sin \theta & \cos\theta \end{pmatrix}$. It then follows that $$\begin{pmatrix} c_1 & 0 \\ d_1 & d_2 \end{pmatrix} = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix} \begin{pmatrix} \cos\theta & \sin \theta \\ - \sin \theta & \cos\theta \end{pmatrix} $$ We know that the product of determinants equals the determinant of the product, and we also know that the determinant of that cosine--sine matrix is equal to $\cos^2\theta+\sin^2\theta=1$. Therefore, $$\begin{vmatrix} c_1 & 0 \\ d_1 & d_2 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} $$
Regarding the parallelogram spanned by $\overline{O S_1}$ and $\overline{O S_2}$, the formula says this area should be equal to $$\begin{vmatrix} c_1 & 0 \\ d_1 & d_2 \end{vmatrix} = c_1 d_2 $$ This is the "still more basic formula" mentioned earlier: it's just the "base times height" formula for the area of a paralellogram, from elementary geometry. This completes the proof.
