Answer to Question #128416 in Analytic Geometry for Nikhil

Question

Show that the points (2,0,1),(0,-4,3) and (-2,5,0) are non-collinear. Hence, find the equation of plane passing through them

Answer 1

Let check the determinant is non zero or not

$$\begin{vmatrix} 2 & 0&1 \\ 0 & -4&3\\ -2&5&0 \end{vmatrix}=-19\neq0$$

Thus points are non collinear.

Equation of plane:

$$\begin{vmatrix} x-2&y-0&z-1\\ 2-0&0-(-4)&1-3 \\ 0-(-2)&-4-5&3-0 \end{vmatrix}=\begin{vmatrix} x-2&y-0&z-1\\ 2&4&-2 \\ 2&-9&3 \end{vmatrix}=0\\ \implies-6(x-2)-10(y-0)+-26(z-1)=0\\\implies -6x-10y-26z+38=0$$