### Three Dimensional Geometry Question 34

#### Question 34 - 01 February - Shift 2

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 ______.

## Show Answer

#### Answer: 10

#### Solution:

#### Formula: Section Formula, Direction Cosines And Direction Ratios

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

Given line $A B: \frac{x-2}{5}=\frac{y-4}{10}=\frac{z+3}{-4}=\lambda$

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 $(\frac{1}{3}, \frac{2}{3},-\frac{5}{3})$

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

$\overrightarrow{{}CD}=(-\mu+\frac{2}{3}) \hat{i}+(2 \mu-\frac{14}{3}) \hat{j}+(3 \mu-\frac{1}{3}) \hat{k}$

$(-\mu+\frac{2}{3})(-1)+(2 \mu-\frac{14}{3}) 2+(3 \mu-\frac{1}{3}) 3=0$

$\mu=\frac{11}{14}$

$\overrightarrow{{}CD}=\frac{-5}{42}, \frac{-130}{42}, \frac{85}{42}$

Direction ratios $(-1,-26,17)$

$a+b+c=10$