Given :

A ≡ (2, -1, 1)

Plane parallel to the required plane : 3x + 2y – z = 7

To Find : Equation of plane

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

If a plane is passing through point A, then equation of plane is

Where,

Answer :

For point A ≡ (2, -1, 1), position vector is

As required plane is parallel to 3x + 2y – z = 7.

Therefore, normal vector of given plane is also perpendicular to required plane

Now,

= 6 – 2 – 1

= 3

Equation of the plane passing through point A and perpendicular to vector is

As

= 3x + 2y – z

Therefore, equation of the plane is

3x + 2y – z = 3

3x + 2y – z - 3 = 0