Verify that the equation of a plane is this determinant. Verify that the equation of a plane that passes through three points $(a_{1},a_{2},a_{3}),(b_{1},b_{2},b_{3}),(c_{1},c_{2},c_{13})$ of the space is:
$$\begin{vmatrix}
1 & x_{1} & x_{2} & x_{3} \\ 
1 & a_{1} & a_{2} & a_{3} \\ 
1 & b_{1} & b_{2} & b_{3} \\ 
1 & c_{1} & c_{2} & c_{3}  \\
\end{vmatrix}= 0$$
I have tried to reduce the determinant by making zeros, but it continues being so enormous to solve by Sarrus:
$$\begin{vmatrix}
1 & x_{1} & x_{2} & x_{3} \\ 
0 & a_{1}-x_{1} & a_{2}-x_{2} & a_{3}-x_{3} \\ 
0 & b_{1}-x_{1} & b_{2}-x_{2} & b_{3}-x_{3} \\ 
0 & c_{1}-x_{1} & c_{2}-x_{2} & c_{3}-x_{3}  \\
\end{vmatrix}= 0$$
Is there a simple way to prove that?
Thanks in advance.

One approach is as follows. Let $a$ denote the vector $(a_1,a_2,a_3)$ and so-forth.  It suffices to show that $x$ lies in the plane through $a,b,c$ if and only if the system of equations
$$
\begin{cases}
k_1 a + k_2 b + k_3 c = x\\
k_1 + k_2 + k_3 = 1
\end{cases}
$$
has a solution for $k_1,k_2,k_3 \in \Bbb R$.  Indeed, the $x$ of the form $x = k_1 a + k_2 b + k_3 c$ for some $k_i$ with $k_1+k_2+k_3 = 1$ form the affine space generated by $a,b,c$.

Another approach: $x$ lies in the plane through $a,b,c$ if and only if the vectors $x-c,a-c,b-c$ are linearly dependent.
