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 pair wise 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 __________.

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answer is 10.

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Detailed Solution

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

$\vec{C D} = (- μ + \frac{2}{3}) \hat{i} + (2 μ - \frac{14}{3}) \hat{j} + (3 μ - \frac{1}{3}) \hat{k} (- μ + \frac{2}{3}) (- 1) + (2 μ - \frac{14}{3}) 2 + (3 μ - \frac{1}{3}) 3 = 0 μ = \frac{11}{14} \vec{C D} = \frac{- 5}{42}, \frac{- 130}{42}, \frac{85}{42}$