The equations of the plane is given by

A(x-x₁)+B(y-y₁)+C(z-z₁)=0

Where A,B and C are the drs of the normal to the plane.

Putting the first point,

=A(x-1)+B(y+1)+C(z-2)=0 …(1)

Putting the second point in Eqn (1)

=A(2-1)+B(0+1)+C(-1-2)=0

A+B-3C=0 …(a)

Putting the third point in Eqn (1)

=A(0-1)+B(2+1)+C(1-2)=0

= -A+3B-C=0 …(b)

Solving (a) and (b) using cross multiplication method

A+B-3C=0

-A+3B-C=0

Put these in Eqn(1)

=8α(x-1)+4α(y+1)+4α(z-2)=0

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

=2x+2+y+1+z-2=0

2x+y+z+1=0

Now the vector perpendicular to this plane is

Now the unit vector of is given by

((C)