To find the image of the origin with respect to the plane \( P \) passing through the points \( (a, 0, 0) \), \( (0, b, 0) \), and \( (0, 0, c) \), we can follow these steps:
### Step 1: Find the equation of the plane
The plane passes through the three points \( (a, 0, 0) \), \( (0, b, 0) \), and \( (0, 0, c) \). We can use the determinant form to find the equation of the plane.
The general form of the equation of a plane passing through the points \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), and \( (x_3, y_3, z_3) \) is given by:
\[
\begin{vmatrix}
x - x_1 & y - y_1 & z - z_1 \\
x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\
x_3 - x_1 & y_3 - y_1 & z_3 - z_1
\end{vmatrix} = 0
\]
Substituting the points \( (a, 0, 0) \), \( (0, b, 0) \), and \( (0, 0, c) \):
\[
\begin{vmatrix}
x - a & y & z \\
-a & b & 0 \\
-a & 0 & c
\end{vmatrix} = 0
\]
Expanding this determinant, we get:
\[
x(bc) - y(ac) + z(ab) = abc
\]
Thus, the equation of the plane can be simplified to:
\[
\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1
\]
### Step 2: Find the normal vector of the plane
The normal vector \( \mathbf{n} \) to the plane can be derived from the coefficients of \( x \), \( y \), and \( z \) in the plane equation:
\[
\mathbf{n} = (b c, -a c, a b)
\]
### Step 3: Find the coordinates of the image of the origin
Let the image of the origin \( (0, 0, 0) \) be \( Q(x', y', z') \). The line connecting the origin and point \( Q \) must be perpendicular to the plane, meaning it is in the direction of the normal vector.
We can express the coordinates of \( Q \) as:
\[
Q = k \cdot \mathbf{n} = k(bc, -ac, ab)
\]
for some scalar \( k \).
### Step 4: Find the midpoint of the line segment from the origin to point \( Q \)
The midpoint \( M \) of the segment \( O(0, 0, 0) \) and \( Q(x', y', z') \) is given by:
\[
M = \left( \frac{x'}{2}, \frac{y'}{2}, \frac{z'}{2} \right) = \left( \frac{kb c}{2}, \frac{-k a c}{2}, \frac{k a b}{2} \right)
\]
### Step 5: Substitute midpoint into the plane equation
Since \( M \) lies on the plane, we substitute its coordinates into the plane equation:
\[
\frac{\frac{kb c}{2}}{a} + \frac{\frac{-k a c}{2}}{b} + \frac{\frac{k a b}{2}}{c} = 1
\]
This simplifies to:
\[
\frac{kbc}{2a} - \frac{ka^2}{2b} + \frac{kab}{2c} = 1
\]
### Step 6: Solve for \( k \)
Multiplying through by \( 2abc \) to eliminate the denominators gives:
\[
k(b^2c + a^2c - a^2b) = 2abc
\]
Thus,
\[
k = \frac{2abc}{b^2c + a^2c - a^2b}
\]
### Step 7: Find the coordinates of the image \( Q \)
Substituting \( k \) back into the expression for \( Q \):
\[
Q = \left( \frac{2abc}{b^2 + a^2 - ab}(bc), \frac{2abc}{b^2 + a^2 - ab}(-ac), \frac{2abc}{b^2 + a^2 - ab}(ab) \right)
\]
### Final Result
The coordinates of the image of the origin with respect to the plane \( P \) are:
\[
Q = \left( \frac{2b^2c^2}{b^2 + a^2 - ab}, \frac{-2a^2c^2}{b^2 + a^2 - ab}, \frac{2a^2b^2}{b^2 + a^2 - ab} \right)
\]
