Question

Let $P$ be a plane $lx+my+nz=0$ containing the line, $\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}$. If pane $P$ divides the line segments $AB$ joining pionts $A(-3,-6,1)$ and $B(2,4,-3)$ in ratio $k:1$ then the value of $k$ is equal to :

• A. $1.5$
• B. $2$
• C. $4$
• D. $3$

Answer 1

The correct option is B $2$


Lines lies on plane
$-l+2m+3n=0\cdots \left(1\right)$
Point on line $(1,-4,-2)$ lies on plane
$l-4m-2n=0\cdots \left(2\right)$
from $\left(1\right)\&\left(2\right)$
$-2m+n=0\Rightarrow 2m=nl=3n+2m\Rightarrow l=4n\Rightarrow l:m:n::4n:\frac{n}{2}:n\therefore l:m:n::8:1:2$

Now equation of plane is
$8x+y+2z=0$
$R$ divide $AB$ in ratio $k:1$
$R:(\frac{-3+2k}{k+1},\frac{-6+4k}{k+1},\frac{1-3k}{k+1})$ lies on plane
$8\left(\frac{-3+2k}{k+1}\right)+\left(\frac{-6+4k}{k+1}\right)+2\left(\frac{1-3k}{k+1}\right)=0\Rightarrow -24+16k-6+4k+2-6k=0\Rightarrow -28+14k=0\therefore k=2$