Find the equations of the planes that passes through three points.

Solution:

(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3)

$\left|\begin{array}{lll}1 & 1 & -1 \\ 6 & 4 & -5 \\ -4 & -2 & 3\end{array}\right|=(12-10)-(18-20)-(-12+16)$

$=2+2-4$

$=0$

Since A, B, C are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are A (1, 1, 0), B (1, 2, 1), and C (−2, 2, −1).

$\left|\begin{array}{lll}1 & 1 & 0 \\ 1 & 2 & 1 \\ -2 & 2 & -1\end{array}\right|=(-2-2)-(2+2)=-8 \neq 0$

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points, $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$, and $\left(x_{3}, y_{3}, z_{3}\right)$, is

$\left|\begin{array}{lll}x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}x-1 & y-1 & z \\ 0 & 1 & 1 \\ -3 & 1 & -1\end{array}\right|=0$

$\Rightarrow(-2)(x-1)-3(y-1)+3 z=0$

$\Rightarrow-2 x-3 y+3 z+2+3=0$

$\Rightarrow-2 x-3 y+3 z=-5$

$\Rightarrow 2 x+3 y-3 z=5$

This is the Cartesian equation of the required plane.