Question

Find the equation of plane passing through $(2,2,1)$ and $(9,3,6)$ and perpendicular plane $x+3y+3z=8$.

Answer 1

Let the equation of the plane be

$ax+by+c=1$

it passes through $(2,2,1)$ and $(9,3,6)$

$\Rightarrow$ $2a+2b+c=\mathrm{1....}\left(1\right)$

$9a+3b+6c=\mathrm{1......}\left(2\right)$

Perpendicular to $x+3y+3z=8$

$a+3b+3c=\mathrm{0.......}\left(3\right)$

From (2) and (3)

$9a+3b+6c=1$

$\Rightarrow$ $8a+3c+(a+3b+3c)=1$

$\Rightarrow$ $8a+3c=\mathrm{1.....}\left(4\right)$

$6a+6b+3c=3$

$2a+6b+6c=0$

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$4a-3c=\mathrm{3......}\left(5\right)$

$\Rightarrow$ $a=\frac{3}{4}(1+c)$

from (4) and (5)

$8a=-3c=2(3+3c)$

$\Rightarrow$ $1-3c=6+6c$

$\Rightarrow$ $-5=9c$

$\Rightarrow$ $c=-\frac{9}{5}$

$\Rightarrow$ $a=\frac{3}{9}(1+c)=\frac{3}{9}(1-\frac{9}{5})\Rightarrow$ $\frac{3}{5}$

$3b=-(a+3c)\Rightarrow$ $b=\frac{1}{3}[\frac{3}{5}+\frac{9}{5}]=\frac{1}{5}+\frac{3}{5}=\frac{4}{5}$

equation of plane $3x-4y+9z+5=0$