The line $x - 2 y + 4 z + 4 = 0$ , $x + y + z - 8 = 0$ intersects the plane $x - y + 2 z + 1 = 0$ at the point

(3, 2, 3)

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(5, 2, 1)

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(2, 5, 1)

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(3, 4, 1)

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Solution

The correct option is B $(2, 5, 1)$
Given lines
$x - 2 y + 4 z + 4 = 0$ ....(1)
$x + y + z - 8 = 0$ .....(2)
Subtracting (2) from (1), we get
$\Rightarrow y - z = 4$ .....(3)
Given equation of plane $x - y + 2 z + 1 = 0$
Since, the line intersects the plane, so using (3), we get
$\Rightarrow x + z = 3$ ......(4)
Hence, $y = 5$ , $z = 1$ and $x = 2$
Hence, option C is correct.

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Similar questions

Q. Find the intersection of the line

x - 2 y + 4 z + 4 = 0

x + y + z - 8 = 0

with the plane

x - y + 2 z + 1 = 0

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M (1, 1, 1)

and intersecting at right angle to the line of intersection of the planes

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The line x−2y+4z+4=0, x+y+z−8=0 intersects the plane x−y+2z+1=0 at the point

The correct option is B (2,5,1)Given lines x−2y+4z+4=0 ....(1) x+y+z−8=0 .....(2)Subtracting (2) from (1), we get⇒y−z=4 .....(3)Given equation of plane x−y+2z+1=0Since, the line intersects the plane, so using (3), we get ⇒x+z=3 ......(4)Hence, y=5, z=1 and x=2Hence, option C is correct.

The line $x - 2 y + 4 z + 4 = 0$ , $x + y + z - 8 = 0$ intersects the plane $x - y + 2 z + 1 = 0$ at the point