Answer :

False

Given,


Equation of line,




Any point on this line should satisfy the plane equation to be on that plane,


Also any point of this line will have a position vector



Now, Given equation of plane



Putting a, in the above equation



= (2 + λ)(3) + (-3 – λ)(1) + (-1 + 2λ)(-1) + 2


= 6 – 3λ – 3 - λ + 1 – 2λ + 2


= 5 – 6λ ≠ RHS


Hence, Line doesn’t lies in the given plane


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