Question

The point of intersection $C$ of the plane $8 x+y+2 z=0$ 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

Plane : $8 x + y +2 z =0$
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 } \rightarrow(-1,-26,17) $
$la + b + cl =10$