If the position vectors of the point $\mathrm{A}$ and $\mathrm{B}$ are $3 \hat{i}+\hat{j}+2 \hat{k}$ and…
If the position vectors of the point $\mathrm{A}$ and $\mathrm{B}$ are $3 \hat{i}+\hat{j}+2 \hat{k}$ and $\hat{i}-2 \hat{j}-4 \hat{k}$ respectively, then the equation of the plane through $\mathrm{B}$ and perpendicular to $\mathrm{AB}$ is
$2 x+3 y+6 z+9=0$
$2 x+3 y+6 z-11=0$
$2 x+3 y+6 z+28=0$
$2 x-3 y-6 z-32=0$
Solution
The normal to the plane is along $\overrightarrow{\mathrm{AB}}$
$\begin{aligned} & =-(\hat{i}-2 \hat{j}-4 \hat{k})-(3 \hat{i}+\hat{j}+2 \hat{k}) \\ & =-2 \hat{i}-3 \hat{j}-6 \hat{k}\end{aligned}$
Hence, d.r.s of normal to the plane are $<2,3,6>$
so equation of the plane
$2 x+3 y+6 z+\lambda=0$
but it passes through $(1,-2,-4)$, so $\lambda=28$
$\Rightarrow 2 x+3 y+6 z+28=0$