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Q.Find the equation of the plane through the line of intersection of the planes x+y+z=6x+y+z=6 and 2x+3y+4z+5=02x+3y+4z+5=0 and passing through the point (1,1,1)(1,1,1). OR Find the image of the point (1,6,3)(1, 6, 3) in the line x1=y−12=z−23\dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2021Subjective· 4mImportance★★★★★
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Take the pencil of planes through the intersection line, impose the point to fix λ\lambda, and simplify.

Any plane through the line of intersection is

(x+y+z−6)+λ(2x+3y+4z+5)=0.(x+y+z-6)+\lambda(2x+3y+4z+5)=0.

It passes through (1,1,1)(1,1,1):

(1+1+1−6)+λ(2+3+4+5)=0 ⇒ −3+14λ=0 ⇒ λ=314.(1+1+1-6)+\lambda(2+3+4+5)=0\ \Rightarrow\ -3+14\lambda=0\ \Rightarrow\ \lambda=\frac{3}{14}.

Substitute and multiply by 1414:

14(x+y+z−6)+3(2x+3y+4z+5)=0,14(x+y+z-6)+3(2x+3y+4z+5)=0,

14x+14y+14z−84+6x+9y+12z+15=0,14x+14y+14z-84+6x+9y+12z+15=0,

20x+23y+26z−69=0.20x+23y+26z-69=0.

(Check (1,1,1)(1,1,1): 20+23+26−69=020+23+26-69=0.)

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