Question

Find the equation of the plane through the points (1,1,1),(-4,0,2) and (3,-1,0)

Answer 1

$A(1;1;1);B(-4;0;2);;C(3;-1;0);\\\begin{vmatrix} x-x_1 & x_2-x_1 &x_3-x_1 \\ y-y_1 & y_2-y_1 &y_3-y_1\\z-z_1 &z_2-z_1 &z_3-z_1 \end{vmatrix}=0\rightarrow\\\begin{vmatrix} x-1 & -5 &2 \\ y-y_1 & -1&-2\\z-z_1 &1 &-1 \end{vmatrix}=0\rightarrow\\(x-1)\times(-1)^2\times\begin{vmatrix} -1 & -2 \\ 1 & -1 \end{vmatrix}+\\+(y-1)\times(-1)^3\times\begin{bmatrix} -5 & 2 \\ 1 & -1 \end{bmatrix}+\\+(z-1)\times(-1)^4\times\begin{vmatrix} -5 & 2 \\ -1 & -2 \end{vmatrix}=0\rightarrow\\3(x-1)-3(y-1)+12(z-1)=0\rightarrow\\3x-3y+12z-12=0\\ Answer:3x-3y+12z-12=0$