Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Since, direction ratio of normal
Are$\left\{2-\left(-3\right)\right\},\mathrm{}\left(3-1\right),\left(4-2\right)$
=$5,\mathrm{}2,\mathrm{}2$
Then, equation of plane is$5\left(x+1\right)+2\left(y-2\right)+2\left(z-1\right)=0$
$=5x+2y+2z-1=0$
$=\overline{r.}\left(5i+2j+2k\right)=1$
Or Equation of plane passing through
$\overline{a}=-i+2j-k$and the position vector of normal$\overline{n}=5i+2j+2k$
is$\left(\overline{r}-\overline{a}\right).\overline{n}=0,\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\overline{r.}\left(5i+2j+2k\right)=1$