(a) show that the following four points are coplanar :
(i) (4,5,1), (0,-1,-1) ,(3,9,4) and (-4,4,4)
(ii) (0,-1,0), (2 , 1 , -1) , (1,1,1) and (3,3,0).
(b) Show that the four points : (0,-1,-1), (4,5,1), (3,9,4) and (-4,4,4) are coplanar.
Also,find the equation of the plane containing them .

Answer

To solve the problem, we need to show that the given points are coplanar and find the equation of the plane containing them. We will use the concept of vectors and the scalar triple product.

### Part (a): Show that the following four points are coplanar

#### (i) Points: \( A(4, 5, 1) \), \( B(0, -1, -1) \), \( C(3, 9, 4) \), \( D(-4, 4, 4) \)

1. **Find the vectors**:
- Vector \( \vec{AB} = B - A = (0 - 4, -1 - 5, -1 - 1) = (-4, -6, -2) \)
- Vector \( \vec{AC} = C - A = (3 - 4, 9 - 5, 4 - 1) = (-1, 4, 3) \)
- Vector \( \vec{AD} = D - A = (-4 - 4, 4 - 5, 4 - 1) = (-8, -1, 3) \)

2. **Form the matrix with these vectors**:
\[ \begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} \]

3. **Calculate the determinant**:
\[ = -4 \begin{vmatrix} 4 & 3 \\ -1 & 3 \end{vmatrix} + 6 \begin{vmatrix} -1 & 3 \\ -8 & 3 \end{vmatrix} - 2 \begin{vmatrix} -1 & 4 \\ -8 & -1 \end{vmatrix} \]
\[ = -4(12 + 3) + 6(3 + 24) - 2(1 - 32) \]
\[ = -4(15) + 6(27) - 2(-31) \]
\[ = -60 + 162 + 62 = 164 \]

Since the determinant is not zero, the points are **not coplanar**.

#### (ii) Points: \( A(0, -1, 0) \), \( B(2, 1, -1) \), \( C(1, 1, 1) \), \( D(3, 3, 0) \)

1. **Find the vectors**:
- Vector \( \vec{AB} = B - A = (2 - 0, 1 - (-1), -1 - 0) = (2, 2, -1) \)
- Vector \( \vec{AC} = C - A = (1 - 0, 1 - (-1), 1 - 0) = (1, 2, 1) \)
- Vector \( \vec{AD} = D - A = (3 - 0, 3 - (-1), 0 - 0) = (3, 4, 0) \)

2. **Form the matrix with these vectors**:
\[ \begin{vmatrix} 2 & 2 & -1 \\ 1 & 2 & 1 \\ 3 & 4 & 0 \end{vmatrix} \]

3. **Calculate the determinant**:
\[ = 2 \begin{vmatrix} 2 & 1 \\ 4 & 0 \end{vmatrix} - 2 \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} \]
\[ = 2(0 - 4) - 2(0 - 3) - 1(4 - 6) \]
\[ = 2(-4) + 6 + 2 = -8 + 6 + 2 = 0 \]

Since the determinant is zero, the points are **coplanar**.

### Part (b): Show that the four points \( (0,-1,-1), (4,5,1), (3,9,4), (-4,4,4) \) are coplanar and find the equation of the plane.

1. **Using the vectors from part (a)**:
- We already found \( \vec{AB} = (4, 6, 2) \), \( \vec{AC} = (3, 10, 5) \), and \( \vec{AD} = (-4, 5, 5) \).

2. **Form the matrix**:
\[ \begin{vmatrix} 4 & 6 & 2 \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{vmatrix} \]

3. **Calculate the determinant**:
\[ = 4 \begin{vmatrix} 10 & 5 \\ 5 & 5 \end{vmatrix} - 6 \begin{vmatrix} 3 & 5 \\ -4 & 5 \end{vmatrix} + 2 \begin{vmatrix} 3 & 10 \\ -4 & 5 \end{vmatrix} \]
\[ = 4(50 - 25) - 6(15 + 20) + 2(15 + 40) \]
\[ = 4(25) - 6(35) + 2(55) \]
\[ = 100 - 210 + 110 = 0 \]

Since the determinant is zero, the points are coplanar.

4. **Find the equation of the plane**:
- The normal vector \( \vec{n} \) can be obtained from the cross product of \( \vec{AB} \) and \( \vec{AC} \).
- Let \( \vec{AB} = (4, 6, 2) \) and \( \vec{AC} = (3, 10, 5) \):
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 6 & 2 \\ 3 & 10 & 5 \end{vmatrix} \]
\[ = \hat{i}(6 \cdot 5 - 2 \cdot 10) - \hat{j}(4 \cdot 5 - 2 \cdot 3) + \hat{k}(4 \cdot 10 - 6 \cdot 3) \]
\[ = \hat{i}(30 - 20) - \hat{j}(20 - 6) + \hat{k}(40 - 18) \]
\[ = \hat{i}(10) - \hat{j}(14) + \hat{k}(22) \]
\[ = (10, -14, 22) \]

5. **Equation of the plane**:
Using point \( A(0, -1, -1) \):
\[ 10(x - 0) - 14(y + 1) + 22(z + 1) = 0 \]
\[ 10x - 14y - 14 + 22z + 22 = 0 \]
\[ 10x - 14y + 22z + 8 = 0 \]

### Final Answers:
- **(a)** Points \( A(4, 5, 1), B(0, -1, -1), C(3, 9, 4), D(-4, 4, 4) \) are not coplanar. Points \( A(0, -1, 0), B(2, 1, -1), C(1, 1, 1), D(3, 3, 0) \) are coplanar.
- **(b)** Points \( (0,-1,-1), (4,5,1), (3,9,4), (-4,4,4) \) are coplanar. The equation of the plane is:
\[ 10x - 14y + 22z + 8 = 0 \]