Question

Let the equation of the plane through the points $(-2,-2,2),(1,1,1),(1,-1,2)$ be $kx+my+nz+p$. Find $k+m+n+p$

• A. $7$
• B. $0$
• C. $4$
• D. $6$

Answer 1

The correct option is B $0$

The vectors of the line joining $(-2,-2,2)$ and $(1,1,1)$ is $\overrightarrow{n_1}$

$=3\overrightarrow{i}+3\overrightarrow{j}-\overrightarrow{k}$.

The vectors of the line joining $(1,1,1)$ and $(1,-1,2)$ $\overrightarrow{n_2}$

$=2\overrightarrow{i}-\overrightarrow{k}$

Hence, the normal vector of the plane containing the three points is given by

$=\overrightarrow{n_1}\times \overrightarrow{n_2}$

$=(3\overrightarrow{i}+3\overrightarrow{j}-\overrightarrow{k})\times (2i-k)$
$=-\overrightarrow{i}+3\overrightarrow{j}+6\overrightarrow{k}$

Therefore the equation of the plane is

$-x+3y+6z=d$
Since it passes through $(1,1,1)$, hence

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

$\Rightarrow d=8$

Hence, the equation of the plane is

$-x+3y+6x-8=0$

$\therefore k=-1;m=3;n=6;p=-8$

$\Rightarrow k+m+n+p=-1+3+6-8=0$