Let $\mathrm{P}$ be a plane $l_{\mathrm{x}}+\mathrm{my}+\mathrm{nz}=0$ containing the line, $\frac{1-\mathrm{x}}{1}=\frac{\mathrm{y}+4}{2}=\frac{\mathrm{z}+2}{3}$. If plane $\mathrm{P}$ divides the line segment $\mathrm{AB}$ joining points $\mathrm{A}(-3,-6,1)$ and $\mathrm{B}(2,4,-3)$ in ratio $\mathrm{k}: 1$ then the value of $\mathrm{k}$ is equal to :
Correct Option: , 3
Point $\mathrm{C}$ is
$\left(\frac{2 k-3}{k+1}, \frac{4 k-6}{k+1}, \frac{-3 k+1}{k+1}\right)$
$\frac{x-1}{-1}=\frac{y+4}{2}=\frac{z+2}{3}$
Plane $l_{\mathrm{X}}+\mathrm{my}+\mathrm{nz}=0$
$l(-1)+m(2)+n(3)=0$
$-l+2 m+3 n=0$
It also satisfy point $(1,-4,-2) l-4 m-2 n=0$
Solving (1) and (2) $2 m+3 n=4 m+2 n$
$\mathrm{n}=2 \mathrm{~m}$
$l-4 m-4 m=0$
$l=8 \mathrm{~m}$
$\frac{l}{8}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{2}$
$l: \mathrm{m}: \mathrm{n}=8: 1: 2$
Plane is $8 x+y+2 z=0$ It will satisfy point $\mathrm{C}$
$8\left(\frac{2 \mathrm{k}-3}{\mathrm{k}+1}\right)+\left(\frac{4 \mathrm{k}-6}{\mathrm{k}+1}\right)+2\left(\frac{-3 \mathrm{k}+1}{\mathrm{k}+1}\right)=0$
$16 \mathrm{k}-24+4 \mathrm{k}-6-6 \mathrm{k}+2=0$
$14 \mathrm{k}=28 \quad \therefore \quad \mathrm{k}=2$