To solve the problem step by step, we will follow the approach outlined in the video transcript.
### Step 1: Identify the points where the plane intersects the axes
The equation of the plane is given as:
\[ x + y + z = 1 \]
- **Point A (x-intercept)**: When \( y = 0 \) and \( z = 0 \):
\[ x + 0 + 0 = 1 \implies x = 1 \]
So, \( A(1, 0, 0) \).
- **Point B (y-intercept)**: When \( x = 0 \) and \( z = 0 \):
\[ 0 + y + 0 = 1 \implies y = 1 \]
So, \( B(0, 1, 0) \).
- **Point C (z-intercept)**: When \( x = 0 \) and \( y = 0 \):
\[ 0 + 0 + z = 1 \implies z = 1 \]
So, \( C(0, 0, 1) \).
### Step 2: Define the point P
Let the coordinates of point \( P \) be \( (a, b, c) \).
### Step 3: Calculate the distances OP, PA, PB, and PC
Using the distance formula, we find the distances:
- **Distance OP**:
\[
OP = \sqrt{(a - 0)^2 + (b - 0)^2 + (c - 0)^2} = \sqrt{a^2 + b^2 + c^2}
\]
- **Distance PA**:
\[
PA = \sqrt{(a - 1)^2 + (b - 0)^2 + (c - 0)^2} = \sqrt{(a - 1)^2 + b^2 + c^2}
\]
- **Distance PB**:
\[
PB = \sqrt{(a - 0)^2 + (b - 1)^2 + (c - 0)^2} = \sqrt{a^2 + (b - 1)^2 + c^2}
\]
- **Distance PC**:
\[
PC = \sqrt{(a - 0)^2 + (b - 0)^2 + (c - 1)^2} = \sqrt{a^2 + b^2 + (c - 1)^2}
\]
### Step 4: Set the distances equal
According to the problem, we have:
\[ OP = PA = PB = PC = D \]
### Step 5: Create equations from the distances
We can set up the following equations based on the distances being equal:
1. From \( OP \) and \( PA \):
\[
a^2 + b^2 + c^2 = (a - 1)^2 + b^2 + c^2
\]
Simplifying gives:
\[
a^2 = (a - 1)^2 \implies a^2 = a^2 - 2a + 1 \implies 2a = 1 \implies a = \frac{1}{2}
\]
2. From \( OP \) and \( PB \):
\[
a^2 + b^2 + c^2 = a^2 + (b - 1)^2 + c^2
\]
Simplifying gives:
\[
b^2 = (b - 1)^2 \implies b^2 = b^2 - 2b + 1 \implies 2b = 1 \implies b = \frac{1}{2}
\]
3. From \( OP \) and \( PC \):
\[
a^2 + b^2 + c^2 = a^2 + b^2 + (c - 1)^2
\]
Simplifying gives:
\[
c^2 = (c - 1)^2 \implies c^2 = c^2 - 2c + 1 \implies 2c = 1 \implies c = \frac{1}{2}
\]
### Step 6: Determine the coordinates of point P
Thus, the coordinates of point \( P \) are:
\[ P\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) \]
### Step 7: Calculate the distance from point P to the plane
Using the distance formula from a point to a plane:
\[
\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
For the plane \( x + y + z - 1 = 0 \) (where \( A = 1, B = 1, C = 1, D = -1 \)):
\[
\text{Distance} = \frac{|1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} - 1|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|\frac{3}{2} - 1|}{\sqrt{3}} = \frac{\frac{1}{2}}{\sqrt{3}} = \frac{1}{2\sqrt{3}}
\]
### Step 8: Calculate \( 4\sqrt{3}D \)
Now, substituting \( D = \frac{1}{2\sqrt{3}} \):
\[
4\sqrt{3}D = 4\sqrt{3} \cdot \frac{1}{2\sqrt{3}} = 4 \cdot \frac{1}{2} = 2
\]
### Final Answer
Thus, the value of \( 4\sqrt{3}D \) is:
\[
\boxed{2}
\]
