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  • The point of intersection C of the plane 8x + y + 2z = 0 and the line joining the point A(–3,–61) and B(2,–4,–3) divides the line segment AB internally in the ratio <im

The point of intersection C of the plane 8x + y + 2z = 0 and the line joining the point A(–3,–61) and B(2,–4,–3) divides the line segment AB internally in the ratio \mathrm{k}: . \text { If a, b, c }(|\mathrm{a}|,|\mathrm{b}|,|\mathrm{c}|) are coprime are the direction ratios of the perpendicular form the point C on the line \frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3} , then \left | a+b+c \right | is equal to

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Answers (1)

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Plane : 8 x+y+2 z=0

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

\text { Any point on line }(5 \lambda+2,10 \lambda+4,-4 \lambda-3)

Point of intersection of line and plane

\begin{aligned} & 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{\mathrm{x}-1}{-1}=\frac{\mathrm{y}+4}{2}=\frac{z+2}{3}=\mu \end{aligned}

\begin{aligned} & \overline{C D}=\left(-\mu+\frac{2}{3}\right) \hat{\mathrm{i}}+\left(2 \mu-\frac{14}{3}\right) \hat{\mathrm{j}}+\left(3 \mu-\frac{1}{3}\right) \hat{\mathrm{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{C D}=\frac{-5}{42}, \frac{-130}{42}, \frac{85}{42} \end{aligned}

Direction \; ratio \rightarrow(-1,-26,17)

|a+b+c|=10

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