The shortest distance between the $y$ -axis and the line $2 x + 3 y + 5 z + 1 = 3 x + 4 y + 6 z + 2 = 0$ is $\frac{2}{\sqrt{k}}$ , then the value of $^{'} k^{'}$ is

2

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3

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5

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6

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Solution

The correct option is C $5$
The plane containing the given line is
$(2 x + 3 y + 5 z + 1) + λ (3 x + 4 y + 6 z + 2) = 0$
$\Rightarrow (2 + 3 λ) x + (3 + 4 λ) y + (5 + 6 λ) z + (1 + 2 λ) = 0$ ....(1)
This plane is parallel to $y$ -axis
$\Rightarrow 3 + 4 λ = 0$
$\Rightarrow λ = - \frac{3}{4}$
Put this value in (1), we get the equation of plane as
$x - 2 z + 2 = 0$
So, the perpendicular distance of the origin from the plane $x - 2 z + 2 = 0$ is $\frac{2}{\sqrt{5}}$

Suggest Corrections

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2 x + 3 y + 5 z + 1 = 3 x + 4 y + 6 z + 2 = 0

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