You need to check if the line L is parallel to the plane `pi` , hence, you need to test if direction vector of the line, `bar L = <-6,-2,6>` is perpendicular to the normal vector to the plane, `bar n = <3,-3,2>` .
You need to remember that two vectors are orthogonal if the dot product is 0, hence, you need to test if the dot product of the direction vector and normal vector is 0, such that:
`bar L*bar n =<-6,-2,6>*<3,-3,2>`
`bar L*bar n = -6*3 + (-2)*(-3) + 6*2`
`bar L*bar n = -18 + 6 + 12 = 0`
Hence, performing the dot product `bar L*bar n` yields bar L*bar n = 0, hence, the vector bar L is perpendicular to the vector bar n, thus the line `L` is parallel to the plane `pi` .