Question

The point of intersection $C$ of the plane $8x+y+2z=0$ and the line joining the points $A(-3,-6,1)$ and $B(2,4,-3)$ divides the line segment $AB$ internally in the ratio $k:1$. If $a,b,c(|a|,|b|$, $|c|$ are coprime) are the direction ratios of the perpendicular from the point $C$ on the line $\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}$, then $|a+b+c|$ is equal to ___

Answer 1

Answer: 10

Plane : $8x+y+2z=0$
Given line $AB:\frac{x-2}{5}=\frac{y-4}{10}=\frac{z+3}{-4}=\lambda$
Any point on line $(5\lambda +2,10\lambda +4,-4\lambda -3)$
Point of intersection of line and plane
$8(5\lambda +2)+10\lambda +4-8\lambda -6=0$
$\lambda =-\frac{1}{3}$
$C\left(\frac{1}{3},\frac{2}{3},-\frac{5}{3}\right)$
$L:\frac{x-1}{-1}=\frac{y+4}{2}=\frac{z+2}{3}=\mu$

$\overrightarrow{CD}=\left(-\mu +\frac{2}{3}\right)\hat{i}+\left(2\mu -\frac{14}{3}\right)\hat{j}+\left(3\mu -\frac{1}{3}\right)\hat{k}$
$\left(-\mu +\frac{2}{3}\right)(-1)+\left(2\mu -\frac{14}{3}\right)2+\left(3\mu -\frac{1}{3}\right)3=0$
$\mu =\frac{11}{14}$
$\overrightarrow{CD}=\frac{-5}{42},\frac{-130}{42},\frac{85}{42}$
$\text{ Direction ratios }\to (-1,-26,17)$
$la+b+cl=10$