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  • If the equation of the plane passing through the line of intersection of the planes <img alt="2 x-y+z=3,4 x-3 y+5 z+9= 0" src="https://entrancecorner.oncodecogs.com/gif.latex?2%20x-y&plus;

If the equation of the plane passing through the line of intersection of the planes 2 x-y+z=3,4 x-3 y+5 z+9= 0 and parallel to the line \frac{\mathrm{x}+1}{-2}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-2}{5}$ is $\mathrm{ax}+\mathrm{by}+\mathrm{cz}+6=0,then \mathrm{a}+\mathrm{b}+\mathrm{c} is equal to : 

Option: 1

15


Option: 2

14


Option: 3

13


Option: 4

12


Answers (1)

best_answer

Using family of planer

P: P_{1}+\lambda P_{2}=0 \Rightarrow P(2+4 \lambda) x-(1+3 \lambda) y+(1+5 \lambda) z=(3-9 \lambda)
P \: is \: \|$ to $\frac{x+1}{-2}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-2}{5}

Then \; for \: \lambda: \overrightarrow{\mathrm{n}}_{\mathrm{p}} \cdot \overrightarrow{\mathrm{v}}_{\mathrm{L}}=0

-2(2+4 \lambda)-4(1+3 \lambda)+5(1+5 \lambda)=0
-3+5 \lambda=0 \Rightarrow \lambda=\frac{3}{5}

Hence : \mathrm{P}: 22 \mathrm{x}-14 \mathrm{y}+20 \mathrm{z}=-12

P: 11 x-7 y+10 z+6=0
\Rightarrow \mathrm{a}=11
\mathrm{b}=-7
\mathrm{c}=10
\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=14
Ans. 2

Posted by

manish painkra

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