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