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Question

Consider three planesP1:x−y+z=1P2:x+y−z=−1P3:x−3y+3z=2Let L1,L2,L3 be the lines of intersection of the planes P2 and P3, P3 and P1, and P1 and P2, respectively.
STATEMENT-1 : At least two of the lines L1, L2 and L3 are non-parallel.
and
STATEMENT -2 : The three planes do not have a common point.

A
Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1
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B
Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement-1
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C
Statement -1 is True, Statement -2 is False
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D
Statement -1 is False, Statement -2 is True
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Solution

The correct option is D Statement -1 is False, Statement -2 is True
Given three planes are
P1:xy+z=1 ...(1)
P1:x+yz=1 ....(2)
and P1:x3y+3z=2 ...(3)
Solving Eqs. (1) and (2), we have
x=0,z=1+y
which does not satisfy Eq. (3)
As, x3y+3z=03y+3(1+y)=3(2)
Statement II is true.
Nest,since we know that direction ratio's of line of intersection of planes a1x+b1y+c1z+d1=0 and
a2x+b2y+c2z+d2=0
b1c2b2c1,c1a2a1c2,a1b1
Using above result, we get
Direction ratio's of lines L1,L2 and L3 are o,2,2;0,4,4;0,2,2 respectively.
All the three lines L1,L2 and L3 are parallel pairwise.
Statement I is false.

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