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

(a) $\left( {1,1, - 1} \right),\left( {6,4, - 5} \right),\left( { - 4, - 2,3} \right)$

(b) $\left( {1,1,0} \right),\left( {1,2,1} \right),\left( { - 2,2, - 1} \right)$

.: (a) Any plane through $(1,1, - 1)$ is
$a(x - 1) + b(y - 1) + c(z + 1) = 0$ ...(1)
Since the plane passes through the points $(6,4, - 5)$and $( - 4, - 2,3)$

$\therefore$ $a(6 - 1) + b(4 - 1) + c( - 5 + 1) = 0$

and $a( - 4 - 1) + b( - 2 - 1) + c(3 + 1) = 0$

$\Rightarrow$ $5a + 3b - 4c = 0$ ...(2)
and $- 5a - 3b + 4c = 0$ ...(3)
Solving (2) and (3), $\cfrac{a}{{12 - 12}} = \cfrac{b}{{20 - 20}} = \cfrac{c}{{ - 15 + 15}}$

$\Rightarrow$ $\cfrac{a}{0} = \cfrac{b}{0} = \cfrac{c}{0} \Rightarrow a,b,c$ can’t be found.

Since the points are collinear, so an infinite number of planes can be found through the given points.

(b) Any plane through $(1,1,0)$ is
$a(x - 1) + b(y - 1) + cz = 0$ ...(1)

Since the plane passes through the points
$(1,2,1)$ and $( - 2,2, - 1)$

$\therefore$ $a(1 - 1) + b(2 - 1) + c(1) = 0$

and $a( - 2 - 1) + b(2 - 1) + c( - 1) = 0$
$\Rightarrow$ $b + c = 0$ ….(2)
and $- 3a + b - c = 0$ ….(3)

Solving (2) and (3), $\cfrac{a}{{ - 1 - 1}} = \cfrac{b}{{ - 3 + 0}} = \cfrac{c}{{0 + 3}}$

$\Rightarrow$ $\cfrac{a}{{ - 2}} = \cfrac{b}{{ - 3}} = \cfrac{c}{3} = k$ (say), where $k \ne 0$

$\therefore$ $a = - 2k,b = - 3k,c = 3k$
Putting these values of $a,b,c$ in (1),

we get
$- 2k(x - 1) - 3k(y - 1) + 3kz = 0$
$\Rightarrow$ $- 2(x - 1) - 3(y - 1) + 3z = 0$

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

$\Rightarrow$ $2x + 3y - 3z = 5,$which is the required equation.