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Question

Let the projection of the line L:x12=y+21=z23 in the plane 3x+2y+z=5 is L. If the distance of a plane from the origin which contains the lines L and L is p, then the value of 6p2 is

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Solution


Required plane is PLL which also contains a line with d.r's (2,1,3).
Since, required plane is perpendicular to 3x+2y+z=5
Required plane is
∣ ∣x1y+2z2213321∣ ∣=0
xyz1=0
Distance of the plane from origin is 13.

Alternate Solution:
L:x12=y+21=z23
a=(1,2,2), b=(2,1,3)
The required plane will pass through (1,2,2).
Let the equation of the plane be a(x1)+b(y+2)+c(z2)=0
Since, the line is contained in the plane
2ab+3c=0 (1)

Since, required plane contains its image also.
Hence, required plane is perpendicular to the plane 3x+2y+z=5
3a+2b+c=0 (2)
From (1) and (2)
a=b=c
So, the equation of the plane is xyz=1
Distance of the plane from origin is 13.

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