Find the normal vector and the plane passing through the lin

Find the normal vector and the plane passing through the line x = 2t - 1, y = 2 + t, z = 2 and the point (1, 0, 1).

Setting t=0 and t=1 gives two points

P:(-1,2,2)

Q:(1,3,2)

And third point which is given is

R:(1,0,1)

Vector QP=(2,1,0)

Vector RP=(2,-2,-1)

Normal vector can be found by taking cross product of QP and RP

n=QP x RP=(-1,2,-6)

For a normal vector (a,b,c) and given point on plane: (x0,y0,z0)

The equation of plane is

a(x-x0)+b(y-y0)+c(z-z0)=0

SO equation of plane is

-(x-1)+2y-6(z-1)=0

-x+2y-6z=-7