# Find the equations of the planes that passes through three points, $(b)\; (1, 1, 0), (1, 2, 1), (-2, 2, -1)$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Toolbox:
• The Cartesian equation of a plane passing through these non-collinear points are
• $\begin{vmatrix}x-x_1 & y-y_1 & z-z_1\\x_2-x_1 & y_2-y_1 & z_2-z_1\\x_3-x_1 & y_3-y_1 & z_3-z_1\end{vmatrix}=0$
Step 1:
Let the given points be A(1,1,0),B(1,2,1) and $C(-2,2,-1)$.
Let us first examine if the points are collinear.
$\begin{vmatrix}1 & 1& 0\\1 & 2 & 1\\-2 & 2 & -1\end{vmatrix}$
On expanding we get
$1(-2-2)-1(-1+2)+0$
$\Rightarrow -4-1\neq 0$
Hence the points are not collinear.
Step 2:
The Cartesian equation of a plane passing through these non-collinear points are$\begin{vmatrix}x-x_1 & y-y_1 & z-z_1\\x_2-x_1 & y_2-y_1 & z_2-z_1\\x_3-x_1 & y_3-y_1 & z_3-z_1\end{vmatrix}=0$
On substituting the values for $(x_1,y_1,z_1)(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$
$\begin{vmatrix}x-1 & y-1& z-0\\1-1 & 2-1 & 1-0\\-2-1 &2-2 & -1+0\end{vmatrix}=0$
Step 3:
On expanding we get
$(x-1)(-2)-3(y-1)+3z=0$
On simplifying we get
$-2x+1-3y+3+3z=0$
$\Rightarrow -2x-3y+3z=-5$
$2x+3y-3z=5$
This is the required Cartesian equation of the plane.