Show that the plane that passes through the three points $A(a_1,a_2,a_3),B=(b_1,b_2,b_3),$ and $C=(c_1,c_2,c_3)$ consists of the points $P=(x,y,z)$ 
Show that the plane that passes through the three points $A(a_1,a_2,a_3),B=(b_1,b_2,b_3),$ and $C=(c_1,c_2,c_3)$ consists of the points $P=(x,y,z)$

Point P can

given by  
So I am stuck on this problem and I think I should use determinants but I'm not sure what that would prove.
 A: First, expand the given determinant by multilinearity on columns. You get 8 determinants, 4 of which are identically 0. The remaining 4 determinants, when you take out the factor of x,y, or z from 3 of them, leave an equation of the form Ax+By+Cz+D=0. So the given determinantal expression is that of a plane. Second, substitute the coordinates of any of the given points into the given determinantal expression, giving a determinant with a row of 0's and therefore equal to 0. So the plane passes through each of the given points. The equation is unique, up to multiplication by a non-zero constant, provided the points are in "general position.'
A: An acceptable proof depends on your definition of a plane, which you haven’t provided. Here are a couple of possibilities.  
Translating the origin to an arbitrary point $X=(x,y,z)$ maps $A$, $B$ and $C$ to $A'=A-X$, $B'=B-X$ and $C'=C-X$, respectively. The plane defined by these latter points contains the origin iff $X$ is coplanar with $A$, $B$ and $C$. This is equivalent to saying that the span of $A'$, $B'$ and $C'$ is at most two-dimensional, which in turn is equivalent to saying that these three vectors are linearly dependent, so their determinant vanishes. Thus, for every point on the original plane, the identity holds. Going in the other direction, if the determinant vanishes, then $A'$, $B'$ and $C'$ are linearly dependent, so there is some plane through the origin that contains them. This plane might not be unique, though. Translating by $X$, we find that $X$, $A$, $B$ and $C$ are also coplanar. For uniqueness, we want the span of $A'$, $B'$ and $C'$ to be two-dimensional, which is the same as saying that they are not colinear.  
For a different approach, let’s say that you know that a general Cartesian equation of a plane has the form $kx+ly+mz+n=0$. If you plug in the three known points you get a system of three equations in the unknown coefficients that can be written in matrix form as $$\begin{bmatrix}a_1&a_2&a_3&1\\b_1&b_2&b_3&1\\c_1&c_2&c_3&1\end{bmatrix} \begin{bmatrix}k\\l\\m\\n\end{bmatrix} = 0.$$ We want the nullity of the matrix on the left to be $1$ (why?), so its rows must be linearly independent, which will occur when the three points are in “general position,” i.e., they’re not colinear. Now, we add another arbitrary point $(x,y,z)$ on the plane to this matrix, obtaining $$\begin{bmatrix}x&y&z&1\\a_1&a_2&a_3&1\\b_1&b_2&b_3&1\\c_1&c_2&c_3&1\end{bmatrix}.$$ This matrix must also have a nullity of $1$, so its determinant must vanish. Subtract the first row from all of the others to get $$\begin{vmatrix}x&y&z&1\\a_1-x&a_2-y&a_3-z&0\\b_1-x&b_2-y&b_3-z&0\\c_1-x&c_2-y&c_3-z&0\end{vmatrix} = 0.$$ Expanding the determinant along the last column gives you the required formula.  
This way of constructing an equation for a surface (curve in $\mathbb R^2$) via a determinant derived from the coordinates of a set of known points and the resulting system of linear equations in the unknown coefficients is applicable to a wide variety of curves and surfaces. For example, the same method applied to three noncolinear points in $\mathbb R^2$ produces $$\begin{vmatrix}x^2+y^2&x&y&1\\a_1^2+a_2^2&a_1&a_2&1\\b_1^2+b_2^2&b_1&b_2&1\\c_1^2+c_2^2&c_1&c_2&1\end{vmatrix} = 0$$ for an equation of the circle through the three points.
