Consider three planes `P_(1):x-y+z=1`, `P_(2):x+y-z=-1` and
`P_(3):x-3y+3z=2`. Let `L_(1),L_(2),L_(3)` be the lines of intersection of the planes `P_(2)` and `P_(3),P_(3)` and `P_(1),P_(1)` and `P_(2)` respectively.
Statement I Atleast two of the lines `L_(1),L_(2)` and `L_(3)` are non-parallel.
Statement II The three planes do not have a common point.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

D

Explanation

Vectors normal to planes `P_(1),P_(2)` and `P_(3)` are `vecn_(1)=hati-hatj+hatk, vecn_(2)=hati+hatj-hatk` and `vecn_(3)=hati-3hatj+3hatk`. Therefore, lines `L_(1),L_(2)` and `L_(3)` are parallel to vectors `vecb_(1)=vecn_(2)xxvecn_(3),vecb_(2)=vecn_(3)xxvecn_(1)` and `vecb_(3)=vecn_(1)xxvecn_(2)` respectively.
We have `vecb_(1)=-4(hatj+hatk),vecb_(2)=2(hatj+hatk),vecb_(3)=2(hatj+hatk)`
Clearly lines `L_(1),L_(2),L_(3)` are parallel.
So statement -1 is not true.
Consider the sytem of equation
`x-y+z=1`
`x+y-z=-1`
`x-3y+3z=2`
This system can be written as `AX=B`
i.e.`[(1,-1,1),(1,1,-1),(1,-3,3)][(x),(y),(z)]=[(1),(-1),(2)]`
`hArr[(1,-1,1),(0,2,-2),(0,-2,2)][(x),(y),(z)]=[(1),(-2),(1)]` Applying `R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1)`
`hArr[(1,-1,1),(0,2,-2),(0,0,0)][(x),(y),(z)]=[(1),(-2),(-1)]`
Clearly, it is an inconsistent system of equations. So, given planes do not have a common point. Consequently, statement -2 is true.