# Let P be a plane l x+m y+n z=0 containing

Question:

Let P be a plane $l x+m y+n z=0$ 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 $A B$ 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 :

1. 1.5

2. 3

3. 2

4. 4

Correct Option: , 3

Solution:

$\left(\frac{2 k-3}{k+1}, \frac{4 k-6}{k+1}, \frac{-3 k+1}{k+1}\right)$

$\frac{x-1}{-1}=\frac{y+4}{2}=\frac{z+2}{3}$

Plane $l x+m y+n z=0$

$l(-1)+m(2)+n(3)=0$

$-l+2 m+3 n=0$......(1)

It also satisfy point $(1,-4,-2)$

$l-4 \mathrm{~m}-2 \mathrm{n}=0$.......(2)

Solving (1) and (2)

$2 m+3 n=4 m+2 n$

$\mathrm{n}=2 \mathrm{~m}$

$l-4 \mathrm{~m}-4 \mathrm{~m}=0$

$l=8 \mathrm{~m}$

$\frac{l}{8}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{2}$

$l: \mathrm{m}: \mathrm{n}=8: 1: 2$

Plane is $8 x+y+2 z=0$

It will satisfy point $C$

$8\left(\frac{2 k-3}{k+1}\right)+\left(\frac{4 k-6}{k+1}\right)+2\left(\frac{-3 k+1}{k+1}\right)=0$

$16 k-24+4 k-6-6 k+2=0$

$14 k=28 \quad \therefore \quad k=2$