Let `A=(1,1,-1),B=(0,2,1)` be two given points. Also, let P:`x+y+z=0` be a plane.
If `A^(')` and `B^(')` are the feet of perpendicular from A and B, respectively, on the plane 'P' then `A^(')B^(')` equals

Options

A. `sqrt(14)/3`
B. `sqrt(5/3)`
C. `sqrt(3)`
D. `sqrt(2/3)`

Answer

To find the distance \( A'B' \) where \( A' \) and \( B' \) are the feet of the perpendiculars from points \( A \) and \( B \) to the plane \( P: x + y + z = 0 \), we will follow these steps:

### Step 1: Identify the coordinates of points A and B
Given:
- \( A = (1, 1, -1) \)
- \( B = (0, 2, 1) \)

### Step 2: Determine the normal vector of the plane
The equation of the plane is \( x + y + z = 0 \). The normal vector \( \vec{n} \) of the plane can be derived from the coefficients of \( x, y, z \):
- \( \vec{n} = (1, 1, 1) \)

### Step 3: Calculate the foot of the perpendicular from point A to the plane
Using the formula for the foot of the perpendicular from a point \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz + D = 0 \):
\[ \frac{x - x_1}{A} = \frac{y - y_1}{B} = \frac{z - z_1}{C} = -\frac{Ax_1 + By_1 + Cz_1 + D}{A^2 + B^2 + C^2} \]
For point \( A(1, 1, -1) \):
- \( A = 1, B = 1, C = 1, D = 0 \)
- \( x_1 = 1, y_1 = 1, z_1 = -1 \)

Calculating \( Ax_1 + By_1 + Cz_1 + D \):
\[ 1 \cdot 1 + 1 \cdot 1 + 1 \cdot (-1) + 0 = 1 + 1 - 1 + 0 = 1 \]
Now substituting into the formula:
\[ \frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z + 1}{1} = -\frac{1}{3} \]
This gives us:
\[ x - 1 = -\frac{1}{3} \implies x = \frac{2}{3} \]
\[ y - 1 = -\frac{1}{3} \implies y = \frac{2}{3} \]
\[ z + 1 = -\frac{1}{3} \implies z = -\frac{4}{3} \]
Thus, the coordinates of \( A' \) are:
\[ A' = \left( \frac{2}{3}, \frac{2}{3}, -\frac{4}{3} \right) \]

### Step 4: Calculate the foot of the perpendicular from point B to the plane
For point \( B(0, 2, 1) \):
- \( x_1 = 0, y_1 = 2, z_1 = 1 \)

Calculating \( Ax_1 + By_1 + Cz_1 + D \):
\[ 1 \cdot 0 + 1 \cdot 2 + 1 \cdot 1 + 0 = 0 + 2 + 1 + 0 = 3 \]
Substituting into the formula:
\[ \frac{x - 0}{1} = \frac{y - 2}{1} = \frac{z - 1}{1} = -\frac{3}{3} = -1 \]
This gives us:
\[ x = -1 \]
\[ y - 2 = -1 \implies y = 1 \]
\[ z - 1 = -1 \implies z = 0 \]
Thus, the coordinates of \( B' \) are:
\[ B' = (-1, 1, 0) \]

### Step 5: Calculate the distance \( A'B' \)
Using the distance formula:
\[ A'B' = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2} \]
Substituting the coordinates of \( A' \) and \( B' \):
\[ A'B' = \sqrt{\left( \frac{2}{3} - (-1) \right)^2 + \left( \frac{2}{3} - 1 \right)^2 + \left( -\frac{4}{3} - 0 \right)^2} \]
Calculating each term:
\[ = \sqrt{\left( \frac{2}{3} + 1 \right)^2 + \left( \frac{2}{3} - 1 \right)^2 + \left( -\frac{4}{3} \right)^2} \]
\[ = \sqrt{\left( \frac{5}{3} \right)^2 + \left( -\frac{1}{3} \right)^2 + \left( -\frac{4}{3} \right)^2} \]
\[ = \sqrt{\frac{25}{9} + \frac{1}{9} + \frac{16}{9}} = \sqrt{\frac{42}{9}} = \frac{\sqrt{42}}{3} \]

Thus, the distance \( A'B' \) is:
\[ \frac{\sqrt{42}}{3} \]

### Final Answer
The distance \( A'B' \) equals \( \frac{\sqrt{42}}{3} \).