Vertices of a parallelogram taken in order are A( 2,-1,4)B(1,0,-1)C( 1,2,3) and D.
Distance of the point P ( 8, 2,-12) from the plane of the parallelogram is

Options

A. `(4sqrt6)/9`
B. `(32sqrt6)/9`
C. `(16sqrt6)/9`
D. none

Explanation

To find the distance of the point \( P(8, 2, -12) \) from the plane of the parallelogram formed by points \( A(2, -1, 4) \), \( B(1, 0, -1) \), and \( C(1, 2, 3) \), we will follow these steps:

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

The vector \( \overrightarrow{AB} \) is calculated as follows:
\[ \overrightarrow{AB} = B - A = (1 - 2, 0 - (-1), -1 - 4) = (-1, 1, -5) \]

The vector \( \overrightarrow{AC} \) is calculated as follows:
\[ \overrightarrow{AC} = C - A = (1 - 2, 2 - (-1), 3 - 4) = (-1, 3, -1) \]

### Step 2: Find the normal vector \( \overrightarrow{n} \) using the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \)

The cross product is calculated using the determinant:
\[ \overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -5 \\ -1 & 3 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ \overrightarrow{n} = \hat{i} \begin{vmatrix} 1 & -5 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & -5 \\ -1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ -1 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -5 \\ 3 & -1 \end{vmatrix} = (1 \cdot -1) - (-5 \cdot 3) = -1 + 15 = 14 \)
2. \( \begin{vmatrix} -1 & -5 \\ -1 & -1 \end{vmatrix} = (-1 \cdot -1) - (-5 \cdot -1) = 1 - 5 = -4 \)
3. \( \begin{vmatrix} -1 & 1 \\ -1 & 3 \end{vmatrix} = (-1 \cdot 3) - (1 \cdot -1) = -3 + 1 = -2 \)

Thus, we have:
\[ \overrightarrow{n} = 14\hat{i} + 4\hat{j} - 2\hat{k} = (14, 4, -2) \]

### Step 3: Find the unit normal vector \( \hat{n} \)

To find the unit normal vector \( \hat{n} \):
\[ |\overrightarrow{n}| = \sqrt{14^2 + 4^2 + (-2)^2} = \sqrt{196 + 16 + 4} = \sqrt{216} = 6\sqrt{6} \]
\[ \hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \left(\frac{14}{6\sqrt{6}}, \frac{4}{6\sqrt{6}}, \frac{-2}{6\sqrt{6}}\right) = \left(\frac{7}{3\sqrt{6}}, \frac{2}{3\sqrt{6}}, \frac{-1}{3\sqrt{6}}\right) \]

### Step 4: Find the vector \( \overrightarrow{AP} \)

The vector \( \overrightarrow{AP} \) is calculated as:
\[ \overrightarrow{AP} = P - A = (8 - 2, 2 - (-1), -12 - 4) = (6, 3, -16) \]

### Step 5: Calculate the distance from point \( P \) to the plane

The distance \( d \) from point \( P \) to the plane can be calculated using the formula:
\[ d = \frac{|\overrightarrow{AP} \cdot \overrightarrow{n}|}{|\overrightarrow{n}|} \]

Calculating \( \overrightarrow{AP} \cdot \overrightarrow{n} \):
\[ \overrightarrow{AP} \cdot \overrightarrow{n} = (6, 3, -16) \cdot (14, 4, -2) = 6 \cdot 14 + 3 \cdot 4 + (-16) \cdot (-2) = 84 + 12 + 32 = 128 \]

Thus, the distance \( d \) becomes:
\[ d = \frac{|128|}{6\sqrt{6}} = \frac{128}{6\sqrt{6}} = \frac{64}{3\sqrt{6}} \]

### Final Result

The distance of the point \( P(8, 2, -12) \) from the plane of the parallelogram is:
\[ d = \frac{64\sqrt{6}}{18} = \frac{32\sqrt{6}}{9} \]