Find the cartesian equation of plane passing through the points `(1,1,1), (1,-1,1)` and `(-7,-3,-5)`.

Answer

To find the Cartesian equation of the plane passing through the points \( A(1, 1, 1) \), \( B(1, -1, 1) \), and \( C(-7, -3, -5) \), we can follow these steps:

### Step 1: Find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)

1. **Calculate vector \( \overrightarrow{AB} \)**:
\[ \overrightarrow{AB} = B - A = (1, -1, 1) - (1, 1, 1) = (0, -2, 0) \]

2. **Calculate vector \( \overrightarrow{AC} \)**:
\[ \overrightarrow{AC} = C - A = (-7, -3, -5) - (1, 1, 1) = (-8, -4, -6) \]

### Step 2: Find the normal vector \( \overrightarrow{N} \) to the plane

To find the normal vector \( \overrightarrow{N} \), we compute the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \).

Using the determinant method:
\[ \overrightarrow{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 0 \\ -8 & -4 & -6 \end{vmatrix} \]

Calculating the determinant:
\[ \overrightarrow{N} = \hat{i} \begin{vmatrix} -2 & 0 \\ -4 & -6 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 0 \\ -8 & -6 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & -2 \\ -8 & -4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ (-2)(-6) - (0)(-4) = 12 \]
2. For \( \hat{j} \):
\[ (0)(-6) - (0)(-8) = 0 \]
3. For \( \hat{k} \):
\[ (0)(-4) - (-2)(-8) = -16 \]

Thus,
\[ \overrightarrow{N} = 12\hat{i} - 0\hat{j} - 16\hat{k} = (12, 0, -16) \]

### Step 3: Use the normal vector and a point to find the equation of the plane

The equation of the plane can be expressed as:
\[ N_1(x - x_0) + N_2(y - y_0) + N_3(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane (we can use point A) and \( (N_1, N_2, N_3) \) are the components of the normal vector \( \overrightarrow{N} \).

Substituting the values:
\[ 12(x - 1) + 0(y - 1) - 16(z - 1) = 0 \]

Expanding this:
\[ 12x - 12 - 16z + 16 = 0 \]
\[ 12x - 16z + 4 = 0 \]

Rearranging gives us the Cartesian equation of the plane:
\[ 12x - 16z + 4 = 0 \]

### Final Answer:
The Cartesian equation of the plane is:
\[ 12x - 16z + 4 = 0 \]