If the lines L1:x=y=z and L2:x=y2=z3 and a line L3 passing through (1,1,1) forms a triangle of area 6sq. unit then point of intersection of third line L3 with second line L2 CANNOT be

A(1,1,1) lies on x = y = z any point on secondline is (λ,2λ,3λ)cos⁡θ=642, ∴sin⁡θ=642 Area of △OAB=12(OA)(OB)sin⁡θ=6⇒123(14λ)642=6⇒λ=2∴B=(2,4,6)

If the lines L1:x=y=z and L2:x=y2=z3 and a line L3 passing through (1,1,1) forms a triangle of area 6sq. unit then point of intersection of third line L3 with second line L2 CANNOT be

A(1,1,1) lies on x = y = z any point on secondline is (λ,2λ,3λ)cos⁡θ=642, ∴sin⁡θ=642 Area of △OAB=12(OA)(OB)sin⁡θ=6⇒123(14λ)642=6⇒λ=2∴B=(2,4,6)

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If the lines $L_{1} : x = y = z$ and $L_{2} : x = \frac{y}{2} = \frac{z}{3}$ and a line $L_{3}$ passing through (1,1,1) forms a triangle of area $\sqrt{6} sq .$ unit then point of intersection of third line $L_{3}$ with second line $L_{2}$ CANNOT be

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$(\frac{4}{3}, \frac{8}{3}, \frac{12}{3})$

$(1, 2, 3)$

$(2, 4, 6)$

$(3, 6, 9)$

answer is A, B, C.

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Detailed Solution

A(1,1,1) lies on x = y = z any point on second

line is $(λ, 2 λ, 3 λ)$

$\cos θ = \frac{6}{\sqrt{42}}, ∴ \sin θ = \frac{\sqrt{6}}{\sqrt{42}}$

$Area of △ O A B = \frac{1}{2} (O A) (O B) \sin θ = \sqrt{6}$

$\Rightarrow \frac{1}{2} \sqrt{3} (\sqrt{14} λ) \frac{\sqrt{6}}{\sqrt{42}} = \sqrt{6} \Rightarrow λ = 2$

$∴ B = (2, 4, 6)$

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If the lines L1:x=y=z and L2:x=y2=z3 and a line L3 passing through (1,1,1) forms a triangle of area 6sq. unit then point of intersection of third line L3 with second line L2 CANNOT be

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If the lines $L_{1} : x = y = z$ and $L_{2} : x = \frac{y}{2} = \frac{z}{3}$ and a line $L_{3}$ passing through (1,1,1) forms a triangle of area $\sqrt{6} sq .$ unit then point of intersection of third line $L_{3}$ with second line $L_{2}$ CANNOT be