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If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{λ}$ and the plane $x + 2 y + 3 z = 4$ is $\cos^{- 1} (\sqrt{5 / 14})$ then $λ$ =

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Correct option is C

Let cos−1(514)=θ⇒cosθ=514⇒sinθ=314D.r's of given line (a1,b1,c1)=(1,2,λ)D.r's of normal to the given plane (a2,b2,c2)=(1,2,3)sinθ=a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22314=1+4+3λ1+4+λ21+4+9λ=23

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If the angle between the line x=y−12=z−3λ and the plane x+2y+3z=4 is cos−1(5/14) then λ=

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If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{λ}$ and the plane $x + 2 y + 3 z = 4$ is $\cos^{- 1} (\sqrt{5 / 14})$ then $λ$ =