Answer :

Given Points :

A = (0, -1, 0)

B = (2, 1, -1)

C = (1, 1, 1)

D = (3, 3, 0)

To Prove : Points A, B, C & D are coplanar.

To Find : Equation of plane passing through points A, B, C & D.

Formulae :

**1)** Position vectors :

If A is a point having co-ordinates (a_{1}, a_{2}, a_{3}), then its position vector is given by,

**2)** Equation of line

If A and B are two points having position vectors then equation of line passing through two points is given by,

**3)** Cross Product :

If are two vectors

then,

**4)** Dot Product :

If are two vectors

then,

**5)** Coplanarity of two lines :

If two lines are coplanar then

**6)** Equation of plane :

If two lines are coplanar then equation of the plane containing them is

Where,

For given points,

A = (0, -1, 0)

B = (2, 1, -1)

C = (1, 1, 1)

D = (3, 3, 0)

Position vectors are given by,

Equation of line passing through points A & D is

Let,

Where,

&

And equation of line passing through points B & C is

Let,

Where,

&

Now,

Therefore,

= 0 + 6 + 0

= 6

……… eq(1)

And

= 16 – 6 – 4

= 6

……… eq(2)

From eq(1) and eq(2)

Hence lines are coplanar

Therefore, points A, B, C & D are also coplanar.

As lines are coplanar therefore equation of the plane passing through two lines containing four given points is

Now,

= 8x - 6y + 4z

From eq(1)

Therefore, equation of required plane is

8x - 6y + 4z = 6

4x - 3y + 2z = 3

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