# Find the equations of the planes that passes through three points. $(1, 1, -1), (6, 4, -5), (-4, -2, 3)$

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Toolbox:
• If three points are collinear,then there will be many planes that will contain them.
• Cartesian form of the plane containing three non collinear points $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ 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:
The given points are $A(1,1,-1),B(6,4,-5)$ and $C(-4,-2,3)$
First let us examine if $A,B$ and $C$ are collinear points.
We can written as
$\begin{vmatrix}1 & 1 & -1\\6 & 4 & -5\\-4 & -2 & 3\end{vmatrix}$
Step 2:
On expanding the above determinant we get,
$1(12-10)-1(18-20)-1(-12+16)$
$\Rightarrow 1(2)-1(-2)-1(4)$
On simplifying we get,
2+2-4=0
Since the determinant value is 0,the points are collinear.
Hence infinite number of planes pass through the three points.