Let the plane P:8x+α1y+α2z+12=0 be parallel to the line L:x+22=y−33=z+45. If the intercept of P on the y-axis is 1, then the distance between P and L is:

2⋅8+α1⋅3+5⋅α2=0 ⇒3α1+5α2=−16 ...(1)y intercept of plane =1⇒−12α1=1⇒α1=−12⇒α2=4So plane P is 8x−12y+4z+12=02x−3y+z+3=0 Point (−2,3,−4) is on LDistance between P & L is −4−9−4+34+9+1=1414=14 units

Let the plane P:8x+α1y+α2z+12=0 be parallel to the line L:x+22=y−33=z+45. If the intercept of P on the y-axis is 1, then the distance between P and L is:

2⋅8+α1⋅3+5⋅α2=0 ⇒3α1+5α2=−16 ...(1)y intercept of plane =1⇒−12α1=1⇒α1=−12⇒α2=4So plane P is 8x−12y+4z+12=02x−3y+z+3=0 Point (−2,3,−4) is on LDistance between P & L is −4−9−4+34+9+1=1414=14 units

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Let the plane $P : 8 x + α_{1} y + α_{2} z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ . If the intercept of P on the y-axis is 1, then the distance between P and L is:

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$\sqrt{\frac{2}{7}}$

$\sqrt{\frac{7}{2}}$

$\sqrt{\frac{6}{14}}$

$\sqrt{14}$

answer is D.

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Detailed Solution

$2 \cdot 8 + α_{1} \cdot 3 + 5 \cdot α_{2} = 0 \Rightarrow 3 α_{1} + 5 α_{2} = - 16 . . . (1)$
y intercept of plane $= 1 \Rightarrow - \frac{12}{α_{1}} = 1 \Rightarrow α_{1} = - 12 \Rightarrow α_{2} = 4$
So plane P is $8 x - 12 y + 4 z + 12 = 0$
$2 x - 3 y + z + 3 = 0 Point (- 2, 3, - 4)$ is on L
Distance between P & L is $|\frac{- 4 - 9 - 4 + 3}{\sqrt{4 + 9 + 1}}| = \frac{14}{\sqrt{14}} = \sqrt{14} units$